Convert $(-3 \sqrt{2}, -3 \sqrt{2})$ from rectangular coordinates to polar coordinates.

Question

Convert $(-3 \sqrt{2}, -3 \sqrt{2})$ from rectangular coordinates to polar coordinates. (Use radians, with your angle in the interval $[0, 2 \pi)$, and with $r > 0$.)

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I got $\left(6, -\frac{3\pi}{4}\right)$, but that's wrong. Any ideas on where I miscalculated? Thanks!

Answer 1

You have

$$x=r\cos(\theta)\;,\;y=r\sin(\theta)$$

$$\implies r^2=x^2+y^2=18+18=36$$

$$\implies r=6$$

and since $x\neq 0, \tan(\theta)=\frac{y}{x}=1$

$\implies \theta=\frac{\pi}{4}$ or $\frac{\pi}{4}+\pi$

but $\cos(\theta)=\frac{x}{r}=-\frac{\sqrt{2}}{2}$

$$\implies \theta=\frac{5\pi}{4}.$$

Answer 2

It's easier using complex numbers: a complex number with module 1 and equal real and imaginary part is either $\mathrm e^{\tfrac{i\pi}4}$ if its real and imaginary parts are positive, or $\mathrm e^{\tfrac{5i\pi}4}$ if they're negative.

Now the complex number $z=-3\sqrt 2-3\sqrt 2 i$ has real and imaginary parts, and its module is $\sqrt{36}=6$, hence it is $\; 6\mathrm e^{\tfrac{5i\pi}4}$. Its polar coordinates are $$\rho=6,\quad\theta=\frac{5\pi}4.$$