Powers and Roots of Complex numbers

Hello Readers

My today’s post is related to find the nth power and nth root of a given complex number.

Power of a complex number:

Given a complex number z=x+iy, the problem is to find the nth power of z.

To answer it, one can use the powerful tool “Demovire’s theorem”.

For a given z=x+iy, write its polar form by using transformation x=rcosθ and y= r sinθ

so,

z=r(cos θ+isin θ)=re^(iθ)   where  (e^(iθ)  means e to the power of iθ)

z^n=r^n (cos nθ+isin nθ)=r^(n) e^(inθ)

eg.  Given z=1+i, find z^(10)?

Solution. Polar form of z= 1+i is z=(2)^{1/2} (cos π/4 + isin π/4)

So z^(10) = (2)^{10/2} (cos 10π/4 +isin 10π/4 )= (2)^5 (cos (2π + π/2) +isin (2π + π/2))=32i.

Exercise.   Find z^20 for given z=1-i.

Fundamental theorem: 

For a given z_0 ≠ 0, there exist n values of z satisfying the equation z^n=z_0.

Now come to the question,

How to find the roots of the equation z^n=z_0?

Let’s consider polar forms,

z^n=r^n (cos nθ+isin nθ)=r^n e^(inθ) and z_0=r_0(cos θ_0+isin θ_0)=r_0e^(iθ_0)

our equation is z^n=z_0 that implies

r^n   e^(i nθ)=r_0  e^(i θ_0) implies

r^n = r_0   and  nθ =  θ_0 + 2kπ where k=0,1,2,3,…..,n-1 for n distinct roots of equation z^n=z_0.

Note: if k=n then nθ = θ_0 + 2nπ and nθ = θ_0 both gives same value of  θ. (take 2 minutes and think about it if you want to learn mathematics).

Therefore,

r = (r_0)^{1/n} and θ = (θ_0 + 2kπ)/n,  where k=0,1,2,3,…..,n-1 for n distinct roots of equation z^n=z_0.

 

Example: Find the roots of an equation z^3=z_0, where z_0=1-i. 

Solution: the equation is z^3 = 1-i.

where

z^3=r^3 (cos 3θ+isin 3θ)=r^3 e^(i3θ) and

z_0  = 1-i = 2^(1/2)(cos (-π/4 )+isin (π/4 ))= 2^{1/2}e^(-iπ/4 )

implies that r^3  (cos 3θ+isin 3θ) = 2^(1/2)(cos (-π/4 )+isin (π/4 ))

implies r^3=2^(1/2) and 3θ = 2kπ-π/4 , where k=0,1,2.

implies r=2^(1/6) and  θ = (2kπ-π/4)/3  where k=0,1,2.

hence roots are z = r e^{i θ } = 2^(1/6) e^{i(2kπ-π/4)/3 }

when k=0, we have Ist root z1 = 2^(1/6)e^{i(-π/4)/3 }=2^(1/6)( cos (-π/12)+isin (-π/12)).

when k=1, we have 2nd root z2 = 2^(1/6)e^{i(2π-π/4)/3 }=2^(1/6)( cos (7π/12)+isin (7π/12)).

when k=2, we have 3rd root z3 = 2^(1/6)e^{i(4π-π/4)/3 }=2^(1/6)( cos (15π/12)+i sin (15π/12)).

 

You can refer the following notes

 

 

 

 

 

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