OK, so we saw something pretty interesting last time when we multiplied together complex numbers using the modulus argument form.

Remember that for two complex numbers which we will write as $z_1=r_1(\cos\theta_1+\sin\theta_1 i)$ and $z_2=r_2(\cos\theta_2+\sin\theta_2 i)$, where $r_i$ are the moduli, and $\theta_i$ are the arguments of $z_i$. If we multiply them together then we get:

$z_1 z_2=r_1 r_2 (\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2))$

Well, what would happen if the two complex numbers were the same? ie. if we have $z=r(\cos\theta+i\sin\theta)$ and we want $z^2$?

Well, then clearly:

$z^2=r^2(\cos 2\theta+i\sin 2\theta)$.

What if we then multiplied this by $z$ one more time:

$z^3=z^2 z=r^3(\cos (2\theta+\theta)+i\sin(2\theta+\theta))=r^3(\cos 3\theta+i\sin 3\theta)$

hmm, do we already see a pattern emerging? Let’s say that we have a complex number with modulus 1. Complex numbers of the form:

$z=\cos\theta+i\sin\theta$

Are clearly modulus 1. We know that the modulus is the square root of the sum of the squares of the real and imaginary parts of a complex numbers so $|z|=\sqrt{\cos^2\theta+\sin^2\theta}=1$.

ok, so how about if we have $z^n$ where $n$ is an integer?

It turns out that there is a theorem, called de Moivre’s theorem which gives a very simple formula for this.

$(\cos\theta+i\sin\theta)^n=\cos (n\theta)+i\sin(n\theta)$

For integer $n$. This is a particularly beautiful formula and the proof of it is by induction. However, we know that for induction we need a base case, and so when something is true for all integers (positive and negative) we must split it into two parts. First it is proved for $n\ge 0$, then it is proved for $n<0$. You can find the full proof in the resource book, but you should try and work it out for yourselves first. How can we use this formula? Well, let’s say you want to calculate $(3+4i)^{17}$, you first need to calculate the modulus and argument of this complex number. We know that $r=\sqrt{3^2+4^2}=5$ and we have that $\arctan \frac{4}{3}\approx 0.927$ is an argument of this (remember that the argument is not unique. Then we have that:

$(3+4i)^{17}=(5(\cos 0.927+i\sin 0.927))^{17}=5^{17}(\cos(17\times 0.927)+\sin(17\times 0.927))$

However, we can see that $17\times 0.927=15.764>2\pi$ and so because arguments are only unique up to integer addition of factors of $2\pi$ we can subtract from 15.764 as many factors of $2\pi$ as we want until the value of it is between $0$ and $2\pi$. In fact, $15.764-2(2\pi)=3.198$ and so this is a reasonable answer. We can see that the answer to our large power of a complex number is a number whose magnitude is $5^{17}$ and which lies at an angle just a little more than $\pi$ anticlockwise from the positive x-axis ie, it’s just under the negative $Re(z)$ axis. The answer is thus:

$5^{17}\cos(3.198)+5^{17}\sin(3.198)i=-0.998429\times 5^{17}-0.0560261 \times 5^{17}i$

Let’s take an example which is slightly simpler to put into a graph. How about if we have $(1+0.1i)^n$ for different $n$. In the graph below we plot the first 100 $n$. The first one, ie. $(1+0.1i)$ is just above the real axis, and is red, and as $n$ increases the points go from red to green. The lines joining the points to the origin are just for reference. The complex numbers are themselves the points.

 How clear is this post?