Answer to Question #114736 in Algebra for ANJU JAYACHANDRAN

Question

Prove that if a complex number z is an nth root of a real number a , then so
is z .

Answer 1

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

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

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

As for De Moivre's Theorem, complex number

$z=cos\phi+isin\phi$ ,

$z^{1/n}=r^{1/n}(cos\alpha+isin\alpha)$ (I)

where $\alpha=(\phi+2\pi k)/n$ ; $k=1,2,3,...,(n-1)$

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

$z=r(cos\phi+isin\phi)$

To take a general equation as

$\bar{z}=a-ib$

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

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

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

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

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

Then proved.