Answer to Question #103132 in Algebra for ea

Question

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

Answer 1

We know that for every complex number which is in the form of a+ib

now,

for any positive integer n let take nth root in the form of x+iy

so,(a+ib)ⁿ = x+iy

Now use De Moivre's Theorem to extend it

so complex number

$z=cos\theta + isin \theta$

which has n distinct root can be determined by

$\sqrt [n]{z}=\sqrt[n]{r}(cos \alpha+i sin\alpha)---------(i)$

where $\alpha =(\theta +360^{\circ}k)/n,$

we can take k=0,1,2,3,..........,(n-1)

so using the above relation we can get the nth root of the complex number and which is in the form of z bar as $z= r(cosθ+sinθ i)$

Now, take conjugate of general equation as

$\bar{z}=a-ib$

In polar coordinates form it will be $\bar {z} = r(cos \theta -i sin \theta )$

now again using De Moivre's theorem If we find roots in the conjugate case

$\sqrt [n]{\bar{z}}=\sqrt[n]{r}(cos \alpha-i \sin\alpha) -------(ii )$

so from equation (i) and (ii) real parts of the nth part of the equation $\sqrt [n]{\bar{z}}=\sqrt [n]{{z}}$ are the same.

Hence proved.