Answer to Question #102763 in Algebra for aaa

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= r($cos\theta + sin \theta$ i)

which has n distinct root can be determined by $\sqrt [n]{z}=\sqrt[n]{r}(cos \alpha+i sin\alpha)$

where , $(\alpha = \theta +360k)/n)$

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

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

Take conjugate of general equation , as

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

If we find roots in the conjugate case and using De Moivre,s theorem

to expand the below$\sqrt [n]{z}=\sqrt[n]{r}(cos \alpha-i sin\alpha)$

we get in this case also we are getting real part as a which is same as of z

Hence proved