(-3,4) to polar coordinates

Question

I tried to convert $(-3,4)$ to polar coordinates and did the following to calculate it.

\begin{align*} r^2 & =x^2+y^2\\ r^2 & =9+16\\ r& =5 \end{align*}

I got the value of $r$ correct, but the value of theta was wrong.

\begin{align*} \tan(\theta) & = \frac{y}{x}\\ \tan(\theta) & = \frac{4}{-3}\\ \theta & = \arctan\left(-\frac{4}{3}\right)\\ \theta & = -.927 \end{align*}

So then I said the answer is $(5, -.927)$ or $(-5, -.927+\pi)$.

The answer is $(-5,5.356)$.

I don't understand why they got this $r$ or this theta. How is this calculated?

Answer 1

First quadrant

$$\theta=\arctan(\frac yx)$$

second quadrant

$$\theta=\frac{\pi}{2}+\arctan(\frac{-x}{y})$$

third quadrant

$$\theta=\pi+\arctan(\frac yx)$$

fourth quadrant

$$\theta=\frac{3\pi}{2}+\arctan(\frac{x}{-y})$$

Remark

You can use the following identities

$$(\forall X>0)\;\;$$ $$\arctan(X)=\frac{\pi}{2}-\arctan(\frac 1X)$$ and $$(\forall X<0)\;\;$$ $$\arctan(X)=-\frac{\pi}{2}-\arctan(\frac 1X)$$

Answer 2

I guess that $\theta$ is supposed to be in the interval $[0, 2\pi]$. The answer $\theta^*$ you got falls out of this interval, so you just need to find its version $\theta^*+2\pi k$ that happens to be in that interval for some integer $k$. In your case quite simply $k=1$ and $2\pi - .927$ gives you the desired answer.