Complex Numbers, Part 4

Now let’s do some maths with complex numbers. First, notice what happens with successive powers of i:

i0 = 1 (anything to the 0 power is 1)
i1 = i
i2 = -1 (by definition)
i3 = i2 × i = –i
i4 = i2 × i2 = 1

If you keep increasing the power, you will keep getting the pattern:
1, i, -1, –i. For higher powers, you can find where you are in the pattern by dividing the power by 4 and looking at the remainder. If the remainder is 0, then the answer is 1. If the remainder is 1, the answer is i. A remainder of 2 gives -1, and a remainder of 3 gives –i. For example, i69 , 69÷4 = 17 with a remainder of 1. So i69 = i.

Now lets’ convert between rectangular and polar forms.

Convert the following to polar form:

\[\begin{array}{l}
{z}_{1} = 1 \,- \sqrt{3}i\\
{z}_{2} = -1+ \sqrt{3}i\end{array}\]

First let’s find the modulus r. Both of these complex numbers have the same modulus because of the squaring of each term:

\[
{r}_{1} = \sqrt{{1}^{2}+{(-\sqrt{3})}^{2}} = {r}_{2} = \sqrt{{(-1)}^{2} + {(\sqrt{3})}^{2}} = \sqrt{4} = 2\]

Now let’s compute Arg(z) for each. Using my calculator, I get the same answer for both:

\[\begin{array}{l}
{\theta }_{1} = \text{tan}^{-1}{\left(\frac{-\sqrt{3}}{1}\right)} = -\frac{\pi }{3}\\
{\theta }_{2} = \text{tan}^{-1}{\left(\frac{\sqrt{3}}{-1}\right)} = -\frac{\pi }{3}\end{array}\]

The second one is wrong. In the first one, the real part is positive and the imaginary part is negative. This puts the complex number in the fourth quadrant in an Argand plot, so -𝜋/3 is the correct angle. But in the second one, the real part is negative and the imaginary part is positive. This puts the complex number in the second quadrant, so clearly -𝜋/3 is wrong:

Your calculator does not know where the minus sign is so it’s just programmed to give an answer in the range -𝜋/2 to 𝜋/2, (-90° to 90°). You however, being smarter than your calculator, know that the minus sign in z2, puts it in the second quadrant. You need to adjust the given angle to put it in the second quadrant as illustrated.

So the polar form of z1 is 2 cis (-𝜋/3). The polar form of z2 is 2 cis 2𝜋/3.

Now let’s go from polar to rectangular. Convert

\[4\sqrt{2}\text{cis} \left (-\frac{3𝜋}{4}
\right )\]

into its rectangular form.

In my last post I gave the equations

\[x = r \text{ cos} 𝜃, y = r \text{ sin} 𝜃\]

So z =4√2cos(-3𝜋/4) + 4√2 sin(-3𝜋/4)i = -4 – 4i. You do not need to worry about quadrant issues converting from polar to rectangular.

In my next post, I will solve some quadratic equations that previously did not have any solutions when we were stuck in the real domain. Now life is more complex.