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Definition, Properties, Formulas, Solved Example Problems - Euler’s Form of the complex number | 12th Mathematics : UNIT 2 : Complex Numbers

Chapter: 12th Mathematics : UNIT 2 : Complex Numbers

Euler’s Form of the complex number

When performing multiplication or finding powers or roots of complex numbers, Euler form can also be used.

Euler’s Form of the complex number

The following identity is known as Euler’s formula

eiθ  = cosθ + i sinθ

Euler formula gives the polar form z = r eiθ

Note

When performing multiplication or finding powers or roots of complex numbers, Euler form can also be used.

 

Example 2.22

Find the modulus and principal argument of the following complex numbers.

(i) √3 + i 

(ii) -√3 + i 

(iii) - √3 - i 

(iv) √3 - i

Solution

(i) √3 + i

Modulus =


Since the complex number has the principal value √3 + i lies in the first quadrant, has the principal value

 θ = α = π/6.

Therefore, the modulus and principal argument of √3 + i are 2 and π/6 respectively.

(ii) -√3 + i

Modulus = 2 and


Since the complex number  -√3 + i lies in the second quadrant has the principal value


Therefore the modulus and principal argument of (ii) -√3 + i are 2 and 5π/6 respectively.

(iii) - √3 - i

r = 2 and α = π /6 .

Since the complex number - √3 - i lies in the third quadrant, has the principal value,


Therefore, the modulus and principal argument of - √3 - i are 2 and -5π/6 respectively.

(iv) √3 - i

r = 2 and α = π/6

Since the complex number lies in the fourth quadrant, has the principal value,

θ = -α = -π/6


Therefore, the modulus and principal argument of

√3 - i  are 2 and - π/6 .

In all the four cases, modulus are equal, but the arguments are depending on the quadrant in which the complex number lies.

 

Example 2.23

Represent the complex number (i) -1- i (ii) 1+ i√3 in polar form.

Solution

 (i) Let -1- i = r(cosθ + isinθ )

We have 


Since the complex number -1- i lies in the third quadrant, it has the principal value,


Note

Depending upon the various values of k , we get various alternative polar forms.

(ii) 1+ i√3


Therefore, the polar form of 1+ i√3 can be written as


 

Example 2.24

Find the principal argument Arg z , when z = -2 / [ 1+i√3 ]. 

Solution


This implies that one of the values of arg z is 2π/3 .

Since 2π/3 lies between -π and π, the principal argument Arg z is 2 π/3.

 

Properties of polar form

Property 1

If z = r (cosθ + i sinθ ), then z-1 =1/r  (cosθ - i sinθ ) .

Proof


 

Property 2

If z1  = r1 (cosθ1  + i sinθ1 ) and  z2  = r2 (cosθ2  + i sinθ2 ) ,

then   z1 z2  = r1r2 (cos (θ1  + θ2 ) + i sin (θ1  + θ2 )) .

Proof

z = r (cosθ1 + i sinθ1) and 

= r (cosθ2  + i sinθ2)


z1 z2   =  r1 (cosθ1  + i sinθ1 ) r2 (cosθ2  + i sinθ2 )

r1r2 ((cosθ1 cosθ2  - sinθ1 sinθ2 ) + i (sinθ1 cosθ2  + sinθ2  cosθ1 ))

z1 z2   =  r1r2 (cos (θ1  + θ2 ) + i sin (θ1  + θ2 )) .

Note

 arg ( z1 z2 ) = θ1 + θ2 = arg ( z1 ) + arg ( z2 ) .

 

Property 3


Proof

Using the polar form of z1 and z2, we have 


 

Example 2.25

Find the product  in rectangular from. 

Solution


 

Example 2.26

Find the quotient  in rectangular form. 

Solution


Which is in rectangular form.

 

Example 2.27

If z = x + iy and , show that x2 + y2 =1. 

Solution


x2 + y2 =1

 

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12th Mathematics : UNIT 2 : Complex Numbers : Euler’s Form of the complex number | Definition, Properties, Formulas, Solved Example Problems

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12th Mathematics : UNIT 2 : Complex Numbers


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