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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.