Answer to Question #114736 in Algebra for ANJU JAYACHANDRAN

Question #114736
Prove that if a complex number z is an nth root of a real number a , then so
is z .
1
Expert's answer
2020-05-19T13:46:54-0400

We can take any complex number in the form of a+iba+ib ,

 take nth root in the form of x+iyx+iy . Here, n is any positive integer.

so, (a+ib)n=x+iy\lparen a+ib \rparen ^{n}=x+iy

As for De Moivre's Theorem, complex number

z=cosϕ+isinϕz=cos\phi+isin\phi ,

z1/n=r1/n(cosα+isinα)z^{1/n}=r^{1/n}(cos\alpha+isin\alpha) (I)

where α=(ϕ+2πk)/n\alpha=(\phi+2\pi k)/n ; k=1,2,3,...,(n1)k=1,2,3,...,(n-1)

so, we can get the  nth root of the complex number as the form of z bar:

z=r(cosϕ+isinϕ)z=r(cos\phi+isin\phi)

To take a general equation as

zˉ=aib\bar{z}=a-ib

Polar coordinate form of this is zˉ=r(cosϕ+isinϕ)\bar{z}=r(cos\phi+isin\phi)

Using De Moivre's theorem again, we get roots in the conjugate case

zˉ1/n=r1/n(cosαisinα)\bar{z}^{1/n}=r^{1/n}(cos\alpha-isin\alpha) (II)

From equations (I) and (II), we can get that real parts of the nth part of the equation

Re(zˉ1/n)=Re(z1/n)Re(\bar{z}^{1/n})=Re(z^{1/n}) are the same.

Then proved.


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