# Powers Of Complex Numbers

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Given $$z,$$ suppose you have to evaluate $${z^n}$$ where $$n$$ is a rational number. How do we evaluate this power is the subject ofthis section.

Let us write $$z$$ in polar (Euler) form:

$z = r{e^{i\theta }}$

Suppose $$n$$ is an integer. Then $${z^n}$$ is   straightforward to evaluate

\begin{align}{l}{z^n} &= {(r{e^{i\theta }})^n}\,\\&= {r^n}{e^{in\theta }}\end{align}

Now, consider the case when $$n$$ is a non-integer rational number. Let us take \begin{align}n = \frac{1}{k}\end{align} where $$k$$ is an integer. You might wonder, what is the problem in following the same approach for \begin{align}n = \frac{1}{k}\end{align} as we did when $$n$$ was an integer:

\begin{align}&{z^n} = {z^{1/k}}\\\,\,\,\,\, &\quad= {(r{e^{i\theta }})^{1/k}}\\\,\,\,\,\, &\quad= {r^{1/k}}{e^{\frac{{i\theta }}{k}}}\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)\end{align}

The problem is that the value that we have obtained in (1) is just one of the values for $${z^n}$$ ; there are other values too. To be precise, in this case, for example $${n^{th}},$$ the power of $$z,$$ which is actually the $${k^{th}}$$ root of $$z,$$ will have $$k$$ different values, the value in (1) being just one of those $$k$$ values.

How? Consider $$z = {e^{i\theta }}$$ and \begin{align}n = \frac{1}{4}\end{align} . How many values does $${z^{1/4}} = {({e^{i\theta }})^{1/4}}$$ have? $${e^{i\theta /4}}$$ will be one of those values. To actually show that we can obtain other values, we write $${e^{i\theta /4}}$$ as:

${e^{i\theta }} = {e^{i(2p\pi + \theta )}} p \in {\Bbb Z}$

This is justified since $${e^{i(2p\pi )}} = 1$$ . Now,

\begin{align}&{({e^{i\theta }})^{1/4}} = {\{ {e^{i(2p\pi + \theta )}}\} ^{1/4}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\quad= {e^{i\left( {\frac{{2p\pi + \theta }}{4}} \right)}}\end{align}

Let us give different integer values to $$p$$ and see what we obtain

\begin{align}&p = 0 \Rightarrow \,\,\,\,z_0^{1/4} = {e^{i\theta /4}}\\&p = 1 \Rightarrow \,\,\,\,z_1^{1/4} = {e^{i\left( {\frac{\pi }{2} + \frac{\theta }{4}} \right)}}\\&p = 2 \Rightarrow \,\,\,\,z_2^{1/4} = {e^{i\left( {\pi + \frac{\theta }{4}} \right)}}\\&p = 3 \Rightarrow z_3^{1/4} = {e^{i\left( {\frac{{3\pi }}{2} + \frac{\theta }{4}} \right)}}\\&p = 4 \Rightarrow z_4^{1/4} = {e^{i\left( {2\pi + \theta /4} \right)}}\, = {e^{i2\pi }} \cdot {e^{i\,\,\theta /4}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\quad\;\;\quad= {e^{i\,\,\theta /4}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\quad\;\;\quad= z_0^{1/4}\end{align}

and so on

Now we give negative integer values to $$p$$

\begin{align}&p = - 1 \Rightarrow z_{ - 1}^{1/4} = {e^{i\left( { - \frac{\pi }{2} + \frac{\theta }{4}} \right)}} = {e^{i\left( { - 2\pi + \frac{{3\pi }}{2} + \frac{\theta }{4}} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\qquad\qquad\qquad\qquad\qquad\quad\;= {e^{ - i2\pi }}{e^{i\left( {\frac{{3\pi }}{2} + \frac{\theta }{4}} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\qquad\quad\;= z_3^{1/4} \end{align}

Similarly,$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,p = - 2 \Rightarrow z_{ - 2}^{1/4} = z_2^{1/4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$

$\ddots$

and so on

We see that there are precisely 4 unique values $${z^{1/4}}$$ of, given by any four consecutive integer values of $$p.$$ For example, we could take $$p$$ from $${0, 1, 2, 3}\;\rm{or}\; {7, 8, 9,10}$$ or whatever you wish. The important thing to remember is that the four values of $$p$$ should be consecutive (the value of the root follows a cycle of 4, $$p$$ and $$p+4$$ will give the same value for the root). Also, this discussion shows that the fourth root of a complex number has four unique values. What about the $${n^{th}}$$ root? $$n$$ different values.

Before proceeding to the $${n^{th}}$$ roots of a complex number, let us encounter De-Moivre’s theorem, which is a formal statement of what we’ve studied up to this point in the current section.

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