3GUsingDeMoivre

Using De Moivre’s Theorem to solve z n = w

Recall that for z = r cis( q ), De Moivre’s theorem states that: math . \qquad z^n=r^n \text{cis} \big( n\theta \big) math

We can use De Moivre to find square roots, cube roots etc of complex numbers.


 * Example 1 **

math \text{Find } z=\sqrt[3]{-2-2i} math


 * Solution:**

Note this is the solution to: math . \qquad z^3=-2-2i math

{Notice this is a cubic so we expect 3 solutions}

First change –2 – 2i to Polar Form

math \\ . \qquad r=\sqrt{(-2)^{2}+(-2)^{2}} \\. \\ . \qquad r =\sqrt{8} = \big( 8 \big)^{\dfrac{1}{2} } = \big( 2^3 \big)^{\dfrac{1}{2} } \\. \\ . \qquad r = 2^{\dfrac{3}{2} } \\.\\ . \qquad \tan \theta =\dfrac{-2}{-2}=1 \; \rightarrow \; \theta =\dfrac{\pi }{4} math

But in 3rd Quadrant, so math . \qquad \theta =\dfrac{5\pi }{4} \qquad \textit{ This is outside Principal Domain but we will fix that later} math

Thus math . \qquad z^3=\sqrt{8} \text{ cis } \left( \dfrac{5\pi }{4}+2k\pi \right), \; k\in \{0,1,2\} math

{The 3 values of k are because we need 3 solutions.}

Take the cube root of both sides math . \qquad z = \Bigg( 2^{\frac{3}{2}} \text{ cis } \Big( \dfrac {5\pi }{4} +2k \pi \Big) \Bigg) ^ {\dfrac{1}{3}}, \; k \in \{0,1,2\} math

math . \qquad z=2^{\frac{1}{2}} \text{cis} \left( \dfrac{5\pi}{12}+\dfrac{2k\pi}{3} \right), \; k\in \{0,1,2\} \quad \text{Using De Moivre's theorem} math

So our three solutions are: math \\ . \qquad k=0 \; \Longrightarrow \;z_1=\sqrt{2} \text{cis} \left( \dfrac{5\pi }{12} \right) \\. \\ . \qquad k=1 \; \Longrightarrow \; z_2=\sqrt{2} \text{cis} \left( \dfrac{13\pi }{12} \right)=\sqrt{2} \text{cis} \left( \dfrac{-11\pi }{12} \right) \\. \\ . \qquad k=2 \; \Longrightarrow \; z_3=\sqrt{2} \text{cis} \left( \dfrac{21\pi }{12} \right) =\sqrt{2} \text{cis} \left( \dfrac{-3\pi }{12} \right)=\sqrt{2} \text{cis}\left( \dfrac{-\pi }{4} \right) math

Note the changes to put the argument back into the Principal Domain of (– p, p ]

On the screen to the right, you can see that the calculator was unable to solve the equation when expressed as a cubic.

When written as a cube root, the calculator found one solution.

When we ask it to write the solution in polar form, we recognise the solution as the third of the three solutions listed above (k = 2).

The lesson here is that we have discovered a limitation of the calculator.

You will have to answer these questions by hand.

NOTE: You may be asked to answer in the form x + yi.

We don't have exact values for these angles, so use decimals

math . \qquad z_1=\sqrt{2} \text{cis} \left( \dfrac{5\pi }{12} \right) math

math \\ . \; \; =\sqrt{2}\left( \cos \left( \dfrac{5\pi }{12} \right)+i\sin \left( \dfrac{5\pi }{12} \right) \right) \\ .\\ . \; \; =\sqrt{2}\left( 0.259+0.966i \right) \\. \\ . \; \;=0.366+1.366i math

etc {Your calculator can give exact values for these using the __**cExpand**__ command (see right)}

Geometric Interpretation of Roots of z n = w

The solutions to z n = w will be equally spaced around the origin.

So for a cube root with 3 solutions, the roots will be on a circle centered at the origin with radius r (the cube root of the modulus) and spaced 120° apart.

math \text{For } z^n = w, \text{ the solutions will be spaced } \dfrac{2\pi}{n} \text{ around the origin} math

Example The solutions to the above example (cube root) can be plotted math . \qquad z=\sqrt[3]{-2-2i} math Can be expressed as math . \qquad z^3=-2-2i math

This has three solutions, equally spaced around a circle with radius: math . \qquad r=\sqrt{2} math

math \\ . \qquad z_1=\sqrt{2} \text{cis} \left( \dfrac{5\pi }{12} \right) \\ .\\ . \qquad z_2=\sqrt{2} \text{cis} \left( \dfrac{-11\pi }{12} \right) \\. \\ . \qquad z_3=\sqrt{2} \text{cis}\left( \dfrac{-\pi }{4} \right) math

{Notice that because the original cubic had a complex number as a coefficient, that the complex roots do NOT appear in conjugate pairs.}

Example

In the diagram below, **A** is a solution to z 4 = w {where w is a complex number}. List the points corresponding to all the solutions.

We expect 4 solutions equally spaced around the circle, so they will be 90° apart.

**A** is one solution, so the other solutions must be in steps of 90° away from **A**

Hence the solutions are at A, E, I, M