2 De Moivre’s theorem and root finding

In this subsection we ask if we can obtain fractional powers of complex numbers; for example what are the values of $8^{1∕3}$ or ${\left(-24\right)}^{1∕4}$ or even ${\left(1+i\right)}^{1∕2}$ ?

More precisely, for these three examples, we are asking for those values of $z$ which satisfy

$z^3-8=0\text{or}{z}^4+24=0\text{or}{z}^2-\left(1+i\right)=0$

Each of these problems involve finding roots of a complex number.

To solve problems such as these we shall need to be more careful with our interpretation of arg $\left(z\right)$ for a given complex number $z$ .

2.1 Arg( $z$ ) revisited

By definition arg $\left(z\right)$ is the angle made by the line representing $z$ with the positive $x$ -axis. See Figure 9(a). However, as the Figure 9(b) shows you can increase $\theta$ by $2\pi$ (or $360^0$ ) and still obtain the same line in the $xy$ plane. In general, as indicated in Figure 9(c) any integer multiple of $2\pi$ can be added to or subtracted from arg $\left(z\right)$ without affecting the Cartesian form of the complex number.

For example:

$z=1+i=\sqrt{2}\left(cos\frac{\pi }{4}+isin\frac{\pi }{4}\right)\text{in polar form}$

However, we could also write, equivalently:

$z=1+i=\sqrt{2}\left(cos\left(\frac{\pi }{4}+2\pi \right)+isin\left(\frac{\pi }{4}+2\pi \right)\right)$

or, in full generality:

$z=1+i=\sqrt{2}\left(cos\left(\frac{\pi }{4}+2k\pi \right)+isin\left(\frac{\pi }{4}+2k\pi \right)\right)k=0,\pm 1,\pm 2,\cdots$

This last expression shows that in the polar form of a complex number the argument of $z$ , arg $\left(z\right)$ , can assume infinitely many different values, each one differing by an integer multiple of $2\pi$ . This is nothing more than a consequence of the well-known properties of the trigonometric functions:

$cos\left(\theta +2k\pi \right)\equiv cos\theta ,sin\left(\theta +2k\pi \right)\equiv sin\theta$ for any integer $k$

We shall now show how we can use this more general interpretation of arg $\left(z\right)$ in the process of finding roots.

Example 8

Find all the values of $8^{1∕3}$ .

Solution

Solving $z=8^{1∕3}$ for $z$ is equivalent to solving the cubic equation $z^3-8=0$ . We expect that there are three possible values of $z$ satisfying this cubic equation. Thus, rearranging: $z^3=8$ . Now write the right-hand side as a complex number in polar form:

$z^3=8\left(cos0+isin0\right)$

(i.e.  $r=\left|8\right|=8$ and arg $\left(8\right)=0$ ). However, if we now generalise our expression for the argument, by adding an arbitrary integer multiple of $2\pi$ , we obtain the modified expression:

$z^3=8\left(cos\left(2k\pi \right)+isin\left(2k\pi \right)\right)k=0,\pm 1,\pm 2,\cdots$

Now take the cube root of both sides:

Now in this expression $k$ can take any integer value or zero. The normal procedure is to take three consecutive values of $k$ (say $k=0,1,2$ ). Any other value of $k$ chosen will lead to a root (a value of $z$ ) which repeats one of the three already determined.

These are the three (complex) values of $8^{\frac{1}{3}}$ . The reader should verify, by direct multiplication, that ${\left(-1+i\sqrt{3}\right)}^3=8$ and that ${\left(-1-i\sqrt{3}\right)}^3=8$ .

The reader may have noticed within this Example a subtle change in notation. When, for example, we write $8^{1∕3}$ then we are expecting three possible values, as calculated above. However, when we write $\sqrt[3]{8}$ then we are only expecting one value: that delivered by your calculator.

Note the two complex roots are complex conjugates (since $z^3-8=0$ is a polynomial equation with real coefficients). In Example 8 we have worked with the polar form. Precisely the same calculation can be carried through using the exponential form of a complex number. We take this opportunity to repeat this calculation but working exclusively in exponential form.

Thus

$z^3=8$

$=8{\text{e}}^{i\left(0\right)}\left(\text{i.e. }r=\left|8\right|=8\text{and}\text{arg}\left(8\right)=0\right)$

$=8{\text{e}}^{i\left(2k\pi \right)}k=0,\pm 1,\pm 2,\cdots$

therefore taking cube roots

$z=\sqrt[3]{8}{\left[{\text{e}}^{i\left(2k\pi \right)}\right]}^{\frac{1}{3}}$

$=\sqrt[3]{8}{\text{e}}^{\frac{i2k\pi }{3}}\text{using De Moivre’s theorem.}$

Again $k$ can take any integer value or zero. Any three consecutive values will give the roots.

These are the three (complex) values of $8^{\frac{1}{3}}$ obtained using the exponential form. Of course at the end of the calculation we have converted back to standard Cartesian form.

Task!

Following the procedure outlined in Example 8 obtain the two complex values of ${\left(1+i\right)}^{1∕2}$ .

Begin by obtaining the polar form (using the general form of the argument) of $\left(1+i\right)$ :

Answer

Now take the square root and use De Moivre’s theorem to complete the solution:

Answer

Exercise

Find all those values of $z$ which satisfy $z^4+1=0$ . Write your values in standard Cartesian form.

Answer

$z_0=\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}{z}_1=-\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}{z}_2=-\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}{z}_3=\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}$