MEMORISE A B AND C D FORMULAS IN TRIGONOMETRY IN AN EASY MANNER

Hello Friends 

welcome to this post of learning trigonometric formulas with me .Most of the Students or Mathematics Learner ,most of the time confuse to remember or memorise A  B and C  D formulas,They  mixed A B and C D formulas with each other and could not reproduce what they have learnt . So today we going to learn new techniques to learn "How to memorise AB and CD formulas" forever. Before this we must have knowledge of different trigonometric values of different angles in different quadrants.

First of all have a quick look at some of  these formulas .

To clear your  all doubts on   " How to Calculate Different Trigonometric values in different quadrants "  in an easy Method. click on the  above  links  .

 

Tricks to Learn    A  B   Formulae  For  sine  angles

When angles are added   i. e  Sin  ( A+B )  

When Angles are added and then their Trigonometric Ratios is taken , and if we have to take the  Sine of  added angles, then it can be done like this.

Start with  sine of angle A  and multiply it with cosine of  angle B and in other part i. e.  After  +ve sign      Start with Cosine of angle A and multiply with Sine of angle B. i.e. start with sine and ends with sine and in middle both the terms are cosine ,and angles start  A then B again A then again B.

Sin (A+B) = Sin A Cos B + Cos A Sin B

When angles are subtracted    i. e  Sin  ( A-B )  

When Angles are subtracted and  their Trigonometric Ratios is taken , and if we have to take the  Sine of  subtracted  angles, then it can be done like this

    
Start with  sine of angle A  and multiply it with cosine of  angle B and in other part i. e.  After  +ve sign    Start with Cosine of angle A and multiply with Sine of angle B. i.e  start with sine and ends with sine and in middle both the terms are cosine ,and angles start  with A then B again A then again B.

Sin (A - B) = Sin A Cos B - Cos A Sin B

Tricks to Learn    A  B   Formulae For  Cosine  angles

When angles are added   i. e  Cos  ( A+B )  

When Angles are added and then their Trigonometric Ratios is taken , and if we have to take the  Cosine  of  added angles, then it can be done like this.

Start with  cosine of angle A  and multiply it with cosine of  angle B and in other part i. e.  After  -ve sign      Start with Sine of angle A and multiply with Sine of angle B. i.e. 1^st   and 2^nd terms are   cosine and  3^rd and 4^th terms    are sine , Angles start with   A then B again A then again B.

"Here  Sum of cosine of  Two angles  is equal to difference of  product of  cosines of both the angles    and product of sine of both the angles ".

Cos (A+B) = Cos A Cos B - Sin A Sin B

When angles are subtracted    i. e  Cos  ( A-B )  

When Angles are subtracted and  then their Trigonometric Ratios is taken , and if we have to take the  cosine of  subtracted  angles, then it can be done like this.

Cos (A - B) = Cos A Cos B + Sin A Sin B

Start with  cosine of  angle A  and multiply it with cosine of  angle B and in other part i. e.  After  +ve sign      Start with Sine of angle A and multiply with Sine of angle B. i. 1^st   and 2^nd terms are   cosine and  3^rd and 4^th  terms    are sine , and angles start with   A then B again A then again B.

"Here  Difference  of cosine of  Two angles  is equal to the  Sum  of  product of  cosines of both the angles    and product of sine of both the angles" .

Want to Learn WHAT IS SET, TYPES OF SETS ,UNION ,INTERSECTION AND VENN DIAGRAMS .

How  to  Memorise     C D   Formulae

To learn C D formulae 

Step 1 

Place 2 for all four formulae and  take Trigonometric Ratio of 1st angle for all four formulae which  is (C+D)/2 and again  trigonometric Ratio of  2nd angle which  is (C-D)/2.

Step 2.1 

For addition of Sine Formula start with sine of 1st angle as mentioned in step 1 and multiply it with cos of  2nd angle as mentioned in step 1.

Step 2.2 

For subtraction of Sine Formula start with cosine of 1st angle as mentioned in step 1 and multiply it with sine of  2nd angle as mentioned in step 1.

Step 3.1

For addition of cosine Formula start with cosine of 1st angle as mentioned in step 1 and multiply it with cosine  of  2nd angle as mentioned in step 1.

Step 3.2

For subtraction of cosine Formula start with sine of 1st angle as mentioned in step 1 and multiply it with sine   of  2nd angle as mentioned in step 1,and do not forget to multiply it with -ve sign.

or  

If you do not want to multiply it with -ve sign  ,then you can change 2nd angle (D-C)/2 instead of (C-D)/2

How to Memorise A   B and C   D  formulas  easily ,watch this video 

Want to learn MATRIX , DIFFERENT TYPES OF MATRICES AND DETERMINANTS .


Want to learn HOW TO SOLVE LINEAR EQUATIONS OF TWO AND THREE VARIABLES || MATRIX METHOD TO SOLVE SYSTEM LINEAR EQUATIONS .

Conclusion

Thanks for devoting your valuable time for the post Easy Tricks to Memorise A B and C D Formulae in Trigonometry and trigonometry's shortcut formulas of this blog. If you found this this blog/post of your concern, Do Follow me on my blog and share this post with your friends . We shall meet again in next post ,till then Good Bye.

Let us learn Multiplication , division , arithmatic and simplification short cut ,tips and  tricks after buying this Book of Magical Mathematics .

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HOW TO MEMORISE DIFFERENT VALUES OF TRIGONOMETRIC ANGLES

Hello Friends welcome to this post of learning trigonometric formulas in an easy way with me . AS most of the Students or Mathematics Learner ,most of the time confuse to remember or memorise value of different trigonometric angles in different quadrants and could not reproduce what they have learnt . So today we going to learn new techniques to learn "How to memorise different values of trigonometric angles in various quadrants" forever. Before this we must have knowledge of different trigonometric values of different angles in different quadrants.

When angle lies in 1st Quadrant

(1) When angle lies in 1st quadrant, then all the t- Ratios have positive values. As in 1st quadrant all the three  sides Perpendicular ,base and hypotenuse of  right angled triangle are positive.

      

(2) When angle involve  π/2 ,then T - Ratios on the left hand side  will changes accordingly , if  t-Ratios start with "co" , then co will be removed and if it is  without "co" then co will be added at the beginning .i.e. sin x will changes to cos x, tan  x to cot x , cos x to sin x ,cosec c to sec x and sec  x to cosec x.

sin (π/2 - x)     =  cos x

cos (π/2 - x)    =   sin x
tan (π/2 - x)     =  cot x
cot (π/2 - x)     =  tan x
sec (π/2 - x)     =  cosec x
cosec (π/2 - x) =  sec x

(3)  When angle involve  π ,then T - Ratios on the left hand side  will NOT changes , put the same  t -ratios which is on the left hand side on right hand side i.e.  sin x  to sin x  , cos x  to cos x  , tan x  to tan  x and so on.

sin (2π + x)      =    sin x
cos (2π + x)    =     cos x
tan (2π + x)      =    tan x
cot (2π + x)     =     cot x
sec (2π + x)     =     sec x
cosec (2π + x) =     cosec x

ALSO READ   MATRIX , DIFFERENT TYPES OF MATRICES AND DETERMINANTS .

When angle lies in 2nd  Quadrant

(1)  when angle lies in 2nd quadrants ,then only two t - ratios sin x and it reciprocal cosec x shall have +ve values and remaining t-Ratios shall have -ve values. As in 2nd  quadrant two  out of the three  sides Perpendicular and  hypotenuse of  right angled triangle are positive  while base is negative. So in all those  t -ratios ,when base involves  they will be negative. So  cos x, tan x ,cot x ,sec x involve with -ve value of base therefore these t- ratios shall be negative.

(2)  When angle involve  π/2 ,then T - Ratios on the left hand side  will changes accordingly , if  t-Ratios start with "co" , then co will be removed and if it is  without "co" then co will be added at the beginning .i.e. sin x will changes to cos x, tan x to cot x and sec x to cosec x.

sin (π/2 + x)     =  sin x

cos (π/2 + x)    = -cos x
tan (π/2 + x)     =- cot x
cot (π/2 + x)     = -tan  x
sec (π/2 + x)     =  -cosec x
cosec (π/2 + x) = sec x

(3)  When angle involve  π ,then T - Ratios on the left hand side  will NOT changes , put the same  t -ratios which is on the left hand side on right hand side i.e.  sin x to sin x , cos x to cos x , tan x to tan x  and so on.

sin (π - x)      =   sin x
cos (π - x)    =     cos x
tan (π - x)      = - tan x
cot (π - x)     =  - cot x
sec (π - x)     =   sec x
cosec (π - x) =  cosec x

ALSO READ EASY TRICKS TO MEMORISE A B AND C D FORMULAS IN TRIGONOMETRY .

When angle lies in 3rd Quadrant

(1)  when angle lies in 3rd quadrants ,then only two t - ratios tan x and it reciprocal cot x shall have +ve values and remaining t-Ratios shall have -ve values. As in 3rd  quadrant two  out of the three  sides Perpendicular and  base of  right angled triangle are negative  and hypotenuse is positive. So in all those  t -ratios ,when one of perpendicular or base  involves  they will be negative. So  sin x, cos x , sec x ,cosec x involve with -ve value of base or perpendicular therefore these t- ratios shall be negative. And tan x and cot x involves with both -ve values of perpendicular and base so they are positive in 3rd quadrant.

(2)  When angle involve  π/2 ,then T - Ratios on the left hand side  will changes accordingly , if  t-Ratios start with "co" , then co will be removed and if it is  without "co" then co will be added at the beginning .i.e. sin x will changes to cos x, tan x to cot x and sec x  to cosec x.

sin (3π/2 - x)     = - cos x
cos (3π/2 - x)    = -  sin x
tan (3π/2 - x)     =    cot x
cot (3π/2 - x)     =    tan x
sec (3π/2 - x)     = - cosec x
cosec (3π/2 - x) = - sec x


(3)  When angle involve  π ,then T - Ratios on the left hand side  will NOT changes , put the same  t -ratios which is on the left hand side on right hand side i.e.  sin x to sin x , cos x to cos x , tan x to tan x  and so on.

sin (π + x)      =   -sin x
cos (π + x)    =    - cos x
tan (π + x)      =   tan x
cot (π + x)     =    cot x
sec (π + x)     =  - sec x
cosec (π + x) =  - cosec x

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Let us understand these learning of trigonometric formulae with the help of this video

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When angle lies in 4th Quadrant

(1)  when angle lies in 4th quadrants ,then only two t - ratios cos x and it reciprocal sec x shall have +ve values and remaining t-Ratios shall have -ve values. As in 4th quadrant two out of the three sides Base and hypotenuse of right angled triangle are positive and perpendicular is negative. So in all those t -ratios ,when perpendicular involves they will be negative. So sin x, tan x ,cot x ,cosec x involve with -ve value of perpendicular therefore these t- ratios shall be negative.

(2)  When angle involve  π/2 ,then T - Ratios on the left hand side  will changes accordingly , if  t-Ratios start with "co" , then co will be removed and if it is  without "co" then co will be added at the beginning .i.e. sin x will changes to cos x, tan x to cot x and sec x  to cosec x.

sin (3π/2 + x)     =   cos x
cos (3π/2 + x)    = - sin x
tan (3π/2 + x)     = - tan x
cot (3π/2 + x)     = - cot  x
sec (3π/2 + x)     =  - sec x

cosec (3π/2 + x) =  cosec x

So if we want to calculate sin 300° ,sin 240° and sin 330°  then it can be find out as follows

sin 300° = sin (270° + 30° ) = - cos 30° = -√3/2

sin 330° = sin (360° - 30° )  = - sin 30° = -1/2

sin 240° = sin (270° - 30° ) = - cos 30° = -√3/2

and if we want to calculate cos 300° , cos 240° and   cos 330°  then it can be find out as follows

cos 300° = cos (270° + 30° ) =  sin 30° = 1/2

cos 330° = cos (360° - 30° )  =  cos 30° = √3/2

cos 240° = cos (270° - 30° ) = - sin 30° = 1/2

(3)  When angle involve  π ,then T - Ratios on the left hand side  will NOT changes , put the same  t -ratios which is on the left hand side on right hand side i.e.  sin x to sin x , cos x to cos x , tan x to tan x  and so on.

sin (2π-x)      =   - sin x
cos (2π-x)    =      cos x
tan (2π-x)      =  - tan x
cot (2π-x)     =   - cot x
sec (2π-x)     =     sec x
cosec (2π-x) =  -  cosec x

sin (-x)      =   - sin x
cos (-x)    =      cos x
tan (-x)      =  - tan x
cot (-x)     =   - cot x
sec (-x)     =     sec x
cosec (-x) =  -  cosec x

GENERALISATION OF THE FORMULAE

So when an angle involves integral multiple of π , i,e -3π, -2π, -π, 2π, 3π, 4π then of the T-Ratios will change , But + or - sign can be added at beginning , e. g . sin(nπ + x) may change to + sin x or - sin x ,similarly cos (nπ + x) may change to + or - cos x depending upon the quadrant in which angle lies.


So if we want to find the value of sin 1110° ,then it can be written as sin (3×360° + 30°) = sin 30° = 1/2 . (As the angle is lying in 1st Quadrant )

Similarly if we want to find the value of sin 1050° ,then it can be written as sin (3×360° - 30°) = - sin 30° = - 1/2. (As the angle is lying in 4th Quadrant )


and when an angle involves odd integral multiple of π/2 i.e. (2n+1)π/2 , i . e -7π/2 , -5π/2 , -3π/2 , π/2 , 3π/2 , 5π/2.

Conclusion

Thanks for devoting your valuable time for the post memorising different values of trigonometric angles in different quadrants of this blog. If you found this  blog/post of your concern, Do Follow me on my blog and share  post with your friends . We shall meet again in next post ,till then Good Bye.

ALSO READ HOW TO SOLVE LINEAR EQUATIONS OF TWO AND THREE VARIABLES || MATRIX METHOD TO SOLVE SYSTEM LINEAR EQUATIONS .

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COMPLEX NUMBERS || MULTIPLICATION || MODULUS || INVERSE || SQUARE ROOTS

Let us discuss concept of number, complex numbers, rational,numbers, real numbers, prime numbers ,complex imaginary numbers, factors of a number, complex number , introduction to complex numbers , operations with complex numbers such as addition of complex numbers , subtraction, multiplying complex numbers, conjugate, modulus  polar form and their Square roots of the complex numbers and complex numbers questions and answers .

Complex Numbers

Introduction to complex numbers

Any number of the type a+ib where a,b are Real numbers and i =√ (-1) is called complex number. Where "a" is called real part and "b" is called imaginary  part . The set of complex number is denoted by  C . These numbers are also called  non real numbers .

Complex Numbers Examples  

3,7 + 3i , -2 + 9i , 3i - 2 , 9i,  3 -7i , 2 , 3i, 6 + √5 i , 6 - √5 i ,√5
Here the numbers 2, 3, √5 are called purely Real and the numbers 3i, 9i are called Imaginary numbers .

Algebra of complex numbers

Addition of complex numbers

If   Z1 = a1+ ib1 and  Z2 = a2+ib2  are two complex numbers then their addition can be calculated by adding their real parts and imaginary parts separately. 

 Z1 + Z2= (a1 + ib1 ) + (a2  + ib2 )
Z1 + Z2= (a1 + a2  ) + (ib1 + ib2 )

Z1 + Z2= (a1 + a2  ) + i(b1 + b2 )

Here   (a1 + a2  ) is called Real Part and   (b1 + b2 ) is called Imaginary Part of resultant .

Let Z1 = 3+5i  and Z2  = 9 - 6i then

Z1 + Z2  = (3 + 5i) + (9 - 6i)

Z1 + Z2 = (3 + 9) + (5i - 6i)  

Z1 + Z2  = 12 - i

Here  12  is its Real Part   -1 is its Imaginary Part

More Examples 

 $\displaystyle{\left({6}+{7}{j}\right)}+{\left({3}-{5}{j}\right)}=$(6+7i)
$Z1=6+7iand$ Z2 =3−5i
Z1  +  Z2  = (3

Z1  +  Z2   = (6+3)+(7−5)i

Z1  +  Z2 = 9+2i

If   Z1  +  Z2  = (12+6i) + (-4 - 5i) ,
Z1  +  Z2 = (12−4) + (6i−5i),
Z1  +  Z2= (12−4) + (6−5)i  ,
Z1  +  Z2 = 8+i

If   Z1 =  6+ √5 i and   Z2  = 7 - √4 i

then  Z1  +  Z2  =  ( 6 + √5 i ) + ( 7 - √4 i) 

          Z1  +  Z2 = (6 + 7) + (√5 i - √4 i) 

           Z1  +  Z2  = 13 - (√5 - √4)i

Subtracting Complex Numbers

If   Z1 = a1 + ib1 and  Z2 = a2 + ib2  are two complex numbers then their subtraction can be calculated by subtracting their real parts and imaginary parts from each  separately. 

 Z1 - Z2= (a1 +ib1 ) - (a2  + ib2 )

Z1 - Z2= (a1 - a2  ) + (ib1 - ib2 )

Z1 - Z2  = (a1 - a2  ) + i(b1 - b2 )

Here   (a1 - a2  ) is its Real Part  and   (b1 - b2 )  is its Imaginary Part.

Let Z1 = 8+5i  and Z2  = 4-6i then

Z1  - Z2  = (8+5i) - (4-6i) = (8-4) - (5i-6i)  = 4+i
Z1 - Z2  = (3+5i) - (9-6i) = (3+9) - (5i-6i)  = 12+i

Examples

 $\displaystyle{\left({6}+{7}{j}\right)}+{\left({3}-{5}{j}\right)}=$(6+7i)

$Z1=6+7iand$ Z2 = 3−5i

Z1  -  Z2  = (6+7i) - (3-5i)

Z1  -  Z2 = (6-3)+(7+5)i

Z1  -  Z2  =3+ 12i

If  Z1  = 12+6i ,  Z2  = -4-5i

   Z1  -  Z2 = (12+6i)-(-4−5i)
  Z1  -  Z2 = (12+4)+(6i+5i) =12+11i 

  Z1 -  Z2  = (12+4)+(6+5)i
 Z1  -  Z2 = 18+9i

Multiplying Complex Numbers

If   Z1 = a1 + ib1 and  Z2 = a2 + ib2  are two complex numbers then their Multiplication can be done as  follows

 Z1 Z2 = (a1 +ib1 ) (a2  + ib2 )

Z1  Z2 = a1  (a2  + ib2 )+ib1(a2  + ib2 )

Z1  Z2  = a1  a2  + i a1 b2 +ib1a2  + ib2 ib1  (because  i²  = −1)

Z1  Z2 = a1  a2  + i a1 b2 +ib1a2  +  i² b2 b1     

Z1  Z2 = (a1  a2  - b1b2)  +i(a1 b2 + a2b1 )

Examples

  If    Z1  = 2 + 6i ,  Z2  = 4 - 5i

 Z1Z2 = (2 + 6i)(4−5i)

Z1Z2 =2(4−5i)+ 6i(4−5i)

Z1Z2 = 8-10i+24i−30i²

Z1Z2 = 8-10i + 24i + 30  

Z1Z2 = (8 + 30)+ (-10i + 24i)

 Z1Z2 = (38)+ (14i)

Z1Z2 = 38 +14i

 If    Z1  = -2-i ,  Z2  = 4+5i

 Z1Z2 = (-2-i)(4+5i)
Z1Z2 = -2(4 + 5i) -i(4 + )

 Z1Z2  =  -8-10i-4i−5i²

 Z1Z2 =   -8-10i-4i+5

 Z1Z2 = (-8+5)+ (-10i-4i)

 Z1Z2 = (-3)+ (-14i)

Z1Z2 = -3 -14i

Now=(3+3i)(1-7i)
= 3×1 + 3×(-7i) + 3i×1+ 3i×(-7i)

 = 3 - 21i + 3i -21i²

 = 3 - 21i + 3i +21 (because i² = −1)

 = 24 - 18i

 How  to Simplify Huge Power  of Iota  

In order to reduce huge/big power of iota to smallest power . First divide the power if iota with 4 and then write first term as power of (4 × quotient) of iota and second term as power of remainder of iota as follows . we know that i⁴ = 1 , implies the contribution of 1st term reduces to "1" .

Note : In order to get quotient after division by 4  , it is sufficient to divide the last two digits of the given power by 4.

Therefore it is the second term which will contribute to answer.

Before we proceed further Remember these result   i² = -1 ,          i³ = -i ,       i⁴ = 1.

So these are some Simplification 

i⁴⁵⁰ =   i^{4×112 .} i²  .= 1× (-1) = -1

i⁴⁵¹ =   i^{4×112 .} i³  .= 1× (-i) = -i

 i⁴⁵² =   i^{4×113 .} i⁰  .= 1× 1 = 1

i⁴⁵³ =   i^{4×113 .} i¹  .= 1× i = i

 ^{i3523 =   i4×855} i³ = 1×(-i)= -i

 ^{i9998 =   i4×2449} i² = 1×(-1) = -1

 ^{i9997 =   i4×2449} i¹ = 1×(i) = i

^{i2221 =   i4×555} i¹ = 1×(i) =  i

^{i2222 =   i4×555} i² = 1×(-1) = -1

^{i2223 =   i4×555} i³ = 1×(-i) = -i

Complex Conjugate of complex Number 

If  z = + iy be any complex number then  x - iy is called its Conjugate.  Simply change the sign of   i   to -i.

Examples

Let Z= 5-7i then Conjugate of this complex number is 5+7i

Let  5 2 - 7 3i  be any complex number  then

 Conjugate of this complex number is 
5 2 + 7 3i  and
  Z= 2+ i be the complex number then Conjugate of this complex number is 2 - i

Modulus of a Complex Number

If  z = + iy be any complex number then  Positive square root of sum of the squares of its Real and Imaginary Parts is called its Conjugate.It is denoted by |Z|

Then 

e. g 

Let Z= -3+2i

then | Z | = √(${-3)}^2+({2)}^2$ )  = √ (13)

If Z = -2-i

Then 

     |Z|   $=$ √ ( (${-2)}^2+({-1)}^2$ )

     |Z|   = √5

If Z= 6 - 8i
Then |Z| = √( 36+64   )  
          |Z|= √ (100) 

          |Z|= 10

Division of two Complex Numbers

To divide Z1 With Z2 Multiply and divide the numerator and Denominator by the Conjugate of  Denominator and then write in a + ib form   after simplification.

Example

Divide  2+3i with 4+7i  

4 + 7i

Multiply Numerator And Denominator  by the conjugate of 4+7i 

2 +3i4 +7i×4 - 7i4 - 7i  =  8 -14i + 12i - 21i²16 + 49

=  29 - 2i65

Now change  into a +ib form

=  29 65 - i265

Example

Divide  3+4i with 2+3i

  3+4i2+3i

Multiply Numerator And Denominator by the conjugate of2+3i 

3+4i2+3i × 2-3i2-3i  =  6 -9i +8i -12i²4+9

=  18- i13

Now change  into a +i b form:

=  18 13 - i113

Multiplicative Inverse of complex Number

If Z = x + iy be any complex number then 1/Z is called it Multiplicative Inverse.

Example

If Z = 1 - ithen Z -1= 1/(1-i)
If  Z = 4+3i 
then Z -1= 1/(4+3i)




If  Z = 1-√ 2i
then Z -1= 1/(1-√ 2 i)

Polar Form of a Complex Number

To Convert  Z = x + iy complex number into Polar Form 

Put $x=r\cos \theta$  ....................(1)

$\backslash displaystyle\{y\}=\{r\}\backslash \; \backslash sin\{\backslash theta\}$

 Then Squaring and adding  (1) and (2)
$we\; get$


 
 
 $r^2=x^2+y^2$ and 

upon dividing  (2)equation  by (1) we get

$\displaystyle \tan{\ }\theta=\frac{y}{{x}}$$\displaystyle{x}={r}\ \cos{\theta}$

Then Z = r( cos θ + i sinθ ) is in Polar Form

Example

Convert the complex number 

7 − 5i    into Polar Form , We need to find r and θ.

Put $x=r\cos \theta$  

$\backslash displaystyle\{y\}=\{r\}\backslash \; \backslash sin\{\backslash theta\}$

\displaystyle{r}=\sqrt{{{x}^{2}+{y}^{2}}
​

$\displaystyle=\sqrt{{{7}^{2}+{\left(-{5}\right)}^{2}}}$

$\displaystyle=\sqrt{{{49}+{25}}}$

$\displaystyle=\sqrt{{{74}}}\approx{8.6}$
8

To find θ, we first find the acute angle α 

which is equal to

$\displaystyle={{\tan}^{{-{1}}}{\left(\frac{5}{{7}}\right)}}$

As the Points (7,-5) lies in fourth Quadrant in Argand Plane ,So   $\displaystyle{7}-{5}{j}$ will be  in the fourth quadrant, 

  θ= 360° - α  = 324.5°

So, expressing $\displaystyle{7}-{5}{j}$ in polar form

7-5i = 8.59 ( cos 324.5° + i sin 324.5° )

Example

Convert the complex number $\backslash displaystyle\{7\}-\{5\}\{j\}$

We need to find r and θ.

$\displaystyle{r}=\sqrt{{{}}}$

$\displaystyle=\sqrt{{{7}^{2}+{\left(-{5}\right)}^{2}}}$

       =    1+2=3

   r = √2

 

To find θ, we first find the acute angle α 

which is equal to

$\displaystyle={{\tan}^{{-{1}}}{\left(\frac{5}{{7}}\right)}}$

α = 45°

As the Points (1,-√2) lies in fourth Quadrant in Argand Plane ,So  1−√2i will be  in the fourth quadrant, 

So θ=360 - 45°

θ= 315°

So, expressing 1$-$√2i  in polar form


Z =  √2(cos 315° + i sin 315°) 

Also Watch   HOW TO CONVERT ANY COMPLEX NUMBER TO IT POLAR FORM 

Square root of complex number

Example 1

Find the square root of 8 + 6i

 Let        z² = (x + yi)²  = 8+6i   Squaring  Both Sides

          \    (x² – y²) + (2xy)i     = 8+6i

          Compare real parts and imaginary parts,we have 

       x² – y² = 8   .............................    (1)

        2xy = 6          .............. ...............  (2)

         

 Using   Identity

  ( x² + y² )²  = (   x² - y² )²  - 4 x²  y² 

    ( x² + y² )²  = 8² + 6²

   ( x² + y² )²  =100

     x² + y² = √(8² + 6²) = 10                  

   x² + y² = 10    .............................. (3)

To find the values of  x and y 

 Adding   (1) and (3) 

 we get  x² – y² + x² + y² = 10 +8   

           2 x² = 18 

          x² = 9 

           x = ±3  

Put   x = 3  and x = -3 and    in equation (2)    

we get     y = 1 when x=3  

and          y = -1 when x= -3

From (2) we concluded that  x and y are of same  sign,

 (x = 3 and y = 1)    or   (x = -3 or y =-1)

Therefore required Square roots of   8+6i are

Z = 3+i and Z = -3-i

Example 2

Find the square root of -7+24i

 Let        z² = (x + yi)²  = -7+24i   Squaring  Both Sides

  (x² – y²) + (2xy)i     = -7+24i

  Compare real parts and imaginary parts,we have 

 x² – y² = -7  .............    (1)

  2xy = 24      ..............  (2)

         

 Using   Identity

   ( x² + y² )²  = (   x² - y² )²  - 4 x²  y² 

   ( x² + y² )²  = (-7)² + (24)²

   ( x² + y² )²  = 49 + 576=625

    x² + y² = √(25)                  

   x² + y² = 25    ...................   (3)           

 To find the values of  x and y 

 Adding   (1) and (3) 

 we get         x² – y² + x² + y² = -7 +24   

 2 x² = 18 

  x² = 9 

 x = ±3  

Put   x = 3  and x = -3 and    in equation (2)    

we get     y = 4 when x=3  

and          y = -4when x= -3

  From (2) we concluded that  x and y are of same  sign,

   (x = 3 and y = 4)    or   (x = -3 or y = -4)

Therefore required Square roots of   8 + 6i are

Z= 3+4i and Z = -3-4i

Example 3

Find the square root of -15 - 8i

 Let        z² = (x + yi)²  = -15 - 8i

  Squaring  Both Sides

 (x² – y²) + (2xy)i     = -15 - 8i

 Compare real parts and imaginary parts,we have 

 x² – y² = -15   .............................    (1)

  2xy = -8         .............. ...............  (2)

         

 Using   Identity

  ( x² + y² )²  = (   x² - y² )²  - 4 x²  y² 

  ( x² + y² )²  = -15² + -8²

  ( x² + y² )²  =225+64

   x² + y² = √289                   

 x² + y² = 17    ..............................   (3)           

 To find the values of  x and y 

 Adding   (1) and (3) 

 we get         x² – y² + x² + y² = -15 +17   

 2 x² = 2 

   x² = 1 

  x = ±1 

Put   x = 1  and x = -1 and    in equation (2)    

we get     y =  -4  when x = 1  

and          y = 4 when x = -1

  From (2) we concluded that  x and y are of same  sign,

  (x = 1 and y = -4)    or   (x = -1 or y = 4)

Therefore required Square roots of   -15 - 8i are

Z=1 - 4i  and Z  = -1+ 4i

Example 4

Find the square root of -3 + 4i

 Let        z² = (x + yi)²  =  -3 + 4i

  Squaring  Both Sides
  (x² – y²) + (2xy)i     = -3 + 4i

          Compare real parts and imaginary parts,we have 

                        x² – y² = -3   .............................    (1)

                        2xy =   4         .............. ...............  (2)

         

 Using   Identity

    ( x² + y² )²  = (   x² - y² )²  - 4 x²  y² 

      ( x² + y² )²  = -3² + 4²

       ( x² + y² )²  =9+4

        x² + y² = √25                  

    x² + y² = 5    ..............................   (3)           

 To find the values of  x and y 

  Adding   (1) and (3) 

 we get  x² – y² + x² + y² = -3 +5   

          2 x² = 2 

          x² = 1 

         x = ±1 

Put   x = 1  and x = -1 and    in equation (2)    

we get     y =  2  when x = 1  

and          y = -2 when x = -1

          From (2) we concluded that  x and y are of same  sign,

                  \ (x = 1 and y = 2)    or   (x = -1 or y =-2)

Therefore required Square roots of   -15 - 8i are

Z=1 +2i  and Z  = -1- 2i

HOW TO FIND THE SQUARE ROOTS OF ANY COMPLEX NUMBER IN HINDI

Conclusion

In this post I have discussed Complex number, rational numbers,
real numbers ,prime numbers, modulus ,inverse, polar form,square roots of complex numbers. If this post helped you little bit, then share it with your friends and like this post to boost me to do better, and follow this Blog .We shall meet in next post till then Bye .

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