From Complex number to polar

Question

I have to calculate $(3 + 3i)^{829}$

What I did:

$$|3 + 3i| = \sqrt{3^2 + 3^2} = \sqrt{9 + 9}=\sqrt{18}$$

The argument of $3 + 3i$ is $45º$.

Answer 1

The key is in the title: you need to write the number in polar form: $$x+\mathrm i y = r(\cos\theta+\mathrm i \sin\theta)$$ where $r$ is the modulus of $x+\mathrm i y$ and $\theta$ is the argument.

In your case, you have $x + \mathrm i y = 3 + 3\mathrm i$.

The modulus is given by $\left|3+3\mathrm i\right|=\sqrt{3^2+3^2}=3\sqrt{2}$.

Since $3+3\mathrm i$ lies on the line $y=x$, and is in the upper-right quadrant. The argument is given by $\arg(3+3\mathrm i)=\frac{1}{4}\pi$.

That means that $r=3\sqrt{2}$ and $\theta=\frac{1}{4}\pi$, and hence $$3+3\mathrm i = 3\sqrt{2}\left(\cos(\tfrac{1}{4}\pi)+\mathrm i \sin(\tfrac{1}{4}\pi)\right)$$

De Moivre's Theorem tells is that $(\cos \theta + \mathrm i \sin \theta)^n = \cos (n\theta) + \mathrm i \sin (n\theta)$.

It follows that \begin{eqnarray*} (3+3\mathrm i)^{829} &=& (3\sqrt 2)^{829}\left(\cos(\tfrac{829}{4}\pi)+\mathrm i \sin(\tfrac{829}{4}\pi)\right) \end{eqnarray*}

The sine and cosine functions are $2\pi$-periodic, i.e. they repeat every $2\pi$ and so $\sin(x+2\pi)\equiv \sin x$ just like $\cos(x+2\pi) \equiv \cos x$ for all real $x$. We can subtract multiples of $2\pi=\frac{8}{4}\pi$ from $\frac{829}{4}\pi$ without changing the values of sine and cosine.

Since $829 = 4 \times 206 + 5$ it follows that $\frac{829}{4} = 206 + \frac{5}{4}$ and hence $\frac{829}{4}\pi = 206\pi + \frac{5}{4}\pi$.

Given that sine and cosine are $2\pi$ periodic:

\begin{eqnarray*} \cos(\tfrac{829}{4}\pi) &=& \cos(206\pi + \tfrac{5}{4}\pi) \\ \\ &=& \cos(\tfrac{5}{4}\pi) \\ \\ \sin(\tfrac{829}{4}\pi) &=& \sin(206\pi + \tfrac{5}{4}\pi) \\ \\ &=& \sin(\tfrac{5}{4}\pi) \end{eqnarray*}

It follows that \begin{eqnarray*} (3\sqrt 2)^{829}\left(\cos(\tfrac{829}{4}\pi)+\mathrm i \sin(\tfrac{829}{4}\pi)\right) &=& (3\sqrt 2)^{829}\left(\cos(\tfrac{5}{4}\pi)+\mathrm i \sin(\tfrac{5}{4}\pi)\right) \\ \\ &=& (3\sqrt 2)^{829}\left(-\tfrac{1}{\sqrt 2}-\tfrac{1}{\sqrt 2}\,\mathrm i \right) \\ \\ &=& -3^{829}\times 2^{414} \times (1+\mathrm i) \end{eqnarray*}

This number is massive and shouldn't be written down.

Answer 2

Note that $3+3i=3(1+i)=3\sqrt{2}e^{i\pi/4}=\sqrt{18}e^{i\pi/4}$. Hence $$(3+3i)^{829}=18^{829/2}\cdot e^{i\pi/4\cdot 829}= 18^{829/2}\cdot e^{i\pi/4\cdot (103\cdot 8+5)}= 18^{829/2}\cdot e^{i5\pi/4}$$ because $829=103\cdot 8+5$ and $(e^{i\pi/4})^8=e^{8i\pi/4}=e^{2\pi i}=1$.

P.S. Take a look to this similar exercise: Polar form of Complex numbers