Title: Default for 10.4 De Moivre’s Theorem
Course: HELM Workbooks 2023-2024

De Moivre's Theorem

We have seen that in polar form, if \(z=r(\cos\theta +i\sin\theta)\) and \(w=t(\cos\phi +i\sin\phi )\) then the product \(zw\) is: \[\qquad zw=rt(\cos (\theta +\phi ) +i\sin (\theta +\phi ))\]In particular, if \(r=1, t=1\) and \(\theta =\phi\) (i.e. \(z=w=\cos\theta +i\sin\theta\)), we obtain \[\qquad (\cos\theta +i\sin\theta )^2=\cos 2\theta +i\sin 2\theta\] Multiplying each side of the above equation by \(\cos\theta +i\sin\theta\) gives \[\qquad (\cos\theta +i\sin\theta )^3=(\cos 2\theta +i\sin 2\theta)(\cos\theta +i\sin\theta )=\cos 3\theta +i\sin 3\theta\]on adding the arguments of the terms in the product.

Similarly, \[\qquad (\cos\theta +i\sin\theta )^4=\cos 4\theta +i\sin 4\theta.\]After completing \(p\) such products we have: \[\qquad (\cos\theta +i\sin\theta )^p=\cos p\theta +i\sin p\theta\]where \(p\) is a positive integer.

In fact this result can be shown to be true for those cases in which \(p\) is a negative integer and even when \(p\) is a rational number e.g. \(p= \dfrac12\).

Key Point

If \(p\) is a rational number then \[(\cos\theta + i\sin\theta)^p = \cos(p\theta) + i\sin(p\theta).\] This result is known as De Moivre's Theorem.

Recalling from the exponential form that \(\cos\theta +i\sin\theta =e^{i\theta}\), De Moivre’s Theorem is simply a statement of the laws of indices: \[\qquad (e^{i \theta})^p=e^{i p\theta}.\]

Finding the roots of complex numbers

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

More precisely, for these three examples, we are asking for those values of \(z\) which satisfy \[\qquad z^3-8=0\qquad\text{or}\qquad z^4+24=0\qquad\text{or}\qquad z^2-(1+i)=0\] Each of these problems involve finding roots of a complex number.

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

Revisiting arg\((z)\)

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

[Image]
The argument of a complex number.

Key Point

arg\((z)\) is unique only up to an integer multiple of \(2\pi\).

For example: \[\qquad z=1+i=\sqrt{2}\left(\cos{\pi \over 4}+i\sin{\pi\over 4}\right)\]

in polar form.

However, we could also write: \[\qquad z=1+i=\sqrt{2}\left(\cos\left({\pi \over 4}+2\pi\right)+i\sin\left({\pi\over 4}+2\pi\right)\right)\] or in full generality:\[\qquad z=1+i=\sqrt{2}\left(\cos\left({\pi \over 4}+2k\pi\right)+i\sin\left({\pi\over 4}+2k\pi\right)\right)\qquad k=0,\pm 1,\pm 2,\ldots\]This last expression shows that in the polar form of a complex number, the argument of \(z\), arg\((z)\), can take 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: \[\qquad \cos(\theta +2k\pi )\equiv\cos\theta,\qquad \sin(\theta +2k\pi)\equiv\sin\theta\]for any integer \(k\).

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

Example

Find all the values of \(8^{1/3}\).

Solution

Using polar form

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: \[\qquad z^3=8(\cos 0+i\sin 0)\](i.e. \(r=|8|=8\) and arg(8)=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: \[z^3=8(\cos(2k\pi)+i\sin(2k\pi))\qquad k=0,\pm1,\pm2,\ldots\] Now take the cube root of both sides: \[\begin{aligned} z&=\root 3\of{8}(\cos (2k\pi )+i\sin (2k\pi ))^{1/3}\\ &=\root 3\of{8}\left(\cos\left(2k\pi\over 3\right)+i\sin\left(2k\pi\over 3\right)\right), \end{aligned}\]

by applying De Moivre's theorem.

Now in this expression \(k\) can take any integer value including 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. \[\begin{aligned} \text{So if}\qquad k&=0\qquad z_0=2(\cos 0+i\sin 0)=2\\k&=1\qquad z_1=2\left(\cos {2\pi \over 3}+i\sin {2\pi\over 3}\right)=-1+i\sqrt{3}\\ k&=2\qquad z_2=2\left(\cos {4\pi \over 3}+i\sin {4\pi\over 3}\right)=-1-i\sqrt{3}\end{aligned}\] These are the three (complex) values of \(8^{1/3}\). The reader should verify, by direct multiplication, that \((-1+i\sqrt{3})^3=8\) and that \((-1-i\sqrt{3})^3=8\).

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 \(\root 3\of {8}\) 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).

Using exponential form

Precisely the same calculation can be carried through using the exponential form of a complex number. We take this opportunity to repeat this previous calculation but working exclusively in exponential form.

Thus \[\begin{aligned} z^3 &=8 \\ &=8e^{i (0)}\qquad\text{(i.e. } r=|8|=8\quad \text{and}\quad \text{arg}(8)=0) \\ &=8e^{i (2k\pi )}\qquad k=0,\pm1,\pm2,\ldots\end{aligned}\] therefore taking cube roots \[\begin{aligned} z&=\root 3\of{8}\left[e^{i (2k\pi )}\right]^{1/3} \\ &=\root 3\of{8}e^{i 2k\pi/3} \end{aligned}\]

using De Moivre's theorem.

Again \(k\) can take any integer value or zero. Any three consecutive values will give the roots. \[\begin{aligned} \text{So if}\qquad k&=0\qquad z_0=2e^{i 0}=2 \\ k&=1\qquad z_1=2e^{i 2\pi/3}=-1+i\sqrt{3}\\ k&=2\qquad z_2=2e^{i 4\pi/3}=-1-i\sqrt{3}\end{aligned}\]These are the three (complex) values of \(8^{1/3}\) obtained using the exponential form. Of course at the end of the calculation we have converted back to standard Cartesian form.

Exercise

Use the procedure in the previous example to find the complex values of \(\displaystyle{z = \left([value]\right)^[value]}\).

Begin by obtaining the polar form (using the general form of the argument) of \(z^2\). Write your answer in the form \(r(\cos(\theta+2k\pi) + i\sin(\theta + 2k\pi))\), e.g. \[1+i = \sqrt{2}\left(\cos\left(\frac{\pi}{4}+2k\pi\right) + i\sin\left(\frac{\pi}{4} + 2k\pi\right)\right)\]i.e. do not evaluate cosine and sine.

\(z^2=\quad\)\(\bigg(\)\(\quad+\quad i\) \(\bigg)\)

Now take this to the power \([value]\) and use De Moivre's Theorem. Give your answers in decimal form, rounding to three decimal places where appropriate:

\(z_1=\quad\)\(\quad+\quad i\)  when \(k=0\).

\(z_2=\quad\)\(\quad+\quad i\)  when \(k=1\).
Feedback: We are given \(\displaystyle{z = \left([value]\right)^[value]}.\) Therefore \(z^2 = \displaystyle [value]\). So \(|z^2| = \displaystyle [value]\) and \(\text{arg}(z^2) = \displaystyle [value].\) Therefore you should obtain\[z = [value]\left(\cos\left([value] + 2k\pi\right) + i\sin\left([value] + 2k\pi\right)\right)\] so that\[z_1 = [value] + [value]i;\qquad z_2 = [value] + ([value])i.\]

Exercise

Find all four complex roots of the quartic equation \(z^4 + [value] = 0\).

You should first express \(z^4 = -a\) in the form \(r[\cos((1 + 2k)\pi) + i\sin((1 + 2k)\pi)]\) remembering that \(r\) should be a real number and \(k\in\mathbb{Z}\), then use De Moivre's Theorem to evaluate the fourth roots.

Express your answers in Cartesian form, fully simplified, i.e. in the form \(a+bi.\) Give precise answers.

\(k=0;\qquad z_1=\)

\(k=1;\qquad z_2=\)

\(k=2;\qquad z_3=\)

\(k=3;\qquad z_4=\)
Feedback: We have \[z^4 = -[value] = -[value]e^{2k\pi i} = [value](-1)e^{2k\pi i} = [value]e^{i\pi}e^{2k\pi i} = [value]e^{\pi i(1+2k)}.\] We can now easily express this in polar form:\[z^4 = [value][\cos((1+2k)\pi) + i\sin((1+2k)\pi)].\] Taking this expression to the power \(1/4\) and applying De Moivre's theorem gives us \[z = [value]^{1/4}[\cos((1+2k)\pi) + i\sin((1+2k)\pi)]^{1/4} = [value]^{1/4}\left[\cos\left(\frac{1}{4}(1+2k)\pi\right) + i\sin\left(\frac{1}{4}(1+2k)\pi\right)\right].\] Now we can find the four roots by plugging in the values \(k = 0,1,2,3,\), yielding respectively\[z_1 = [value],\]\[z_2 = [value],\]\[z_3 = [value],\]\[z_4 = [value].\]