Question

How do I find the cube root of a complex number?

Answer 1

Hint: Let $$z=x+iy$$
We know that complex no. is written in the form of $$x+iy$$,so $$z=x+iy$$. Now we will apply De Moivre’s Theorem to find the cube root of a complex number also by using this formula we can find any number of root .we can find its root in any form coordinate as well as polar form .polar form is simple as compared to coordinate form so we will try to solve it in the polar form so firstly we will try to find its nth root and then by putting the value of n we can find its value.

Complete step by step solution:
As we know that de moivre’s theorem is very important in complex number analysis so Now we will apply de moivre’s formula to find cube root or any n root. So the formula is
$\Rightarrow$ $$z^{{\displaystyle \frac{1}{n}}}={\left|z\right|}^{{\displaystyle \frac{1}{n}}}\{\cos ({\displaystyle \frac{\varphi +2k\pi }{n}})+i\sin ({\displaystyle \frac{\varphi +2k\pi }{n}})\}$$
We can also write $$z=r{e}^{i\theta }$$ in polar form so our formula become
$\Rightarrow$ $$(r{e}^{i\theta }{)}^{{\displaystyle \frac{1}{n}}}=(r{)}^{{\displaystyle \frac{1}{n}}}{e}^{i({\displaystyle \frac{\theta +2k\pi }{n}})}fork=0,1,2.....n-1$$
Now putting n=3 we can get its cube root
$$(r{e}^{i\theta }{)}^{{\displaystyle \frac{1}{3}}}=(r^{{\displaystyle \frac{1}{3}}})e^{i({\displaystyle \frac{\theta +2k\pi }{3}})}k=0,1,2$$
We will get all the three roots by putting the values of k in simultaneous manner and hence like this only we can find any nth number of root

Note: We must take care when putting the values of $$\theta$$. As it has cube roots, so the no. of roots which we get will be equal to 3. for the cube root of a complex number. we should get 3 roots in this way we can cross-check our answer. We should try to convert our complex number in polar form to find its root in an easy manner.