Argument of a complex number as arctan

Question

In a calculus book I have it says that the argument $\theta$ of a complex number $a+bi$ can be expressed as $\theta = \arctan{\frac{b}{a}}$ if $a+bi$ is in the first or fourth quadrant or $\theta = \pi + \arctan{\frac{b}{a}}$ if it's in the second or third. I have tried to convince my self that this is true but for some reason I don't end up with the right expression. Can anyone show me a derivation of this?

Answer 1

Note that$$a+bi=\sqrt{a^2+b^2}\left(\frac a{\sqrt{a^2+b^2}}+\frac b{\sqrt{a^2+b^2}}i\right).$$So, if $a+bi=\rho\bigl(\cos(\theta)+\sin(\theta)i\bigr)$, $\cos\theta=\frac a{\sqrt{a^2+b^2}}$ and $\sin\theta=\frac b{\sqrt{a^2+b^2}}$. Therefore$$\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac ba.\tag1$$So:

• If $a+bi$ belongs to the first or the fourth quadrant, then $\theta\in\left(-\frac\pi2,\frac\pi2\right)$, and then $(1)$ tells us that $\theta=\arctan\left(\frac ba\right)$.
• If $a+bi$ belongs to the second or the third quadrant, then $\theta\in\left(\frac\pi2,\frac{3\pi}2\right)$, and then it follows from $(1)$ and from the fact that $\tan$ is periodic with period $\pi$ that $\theta=\pi+\arctan\left(\frac ba\right)$.