Assume I have complex number $z = a + ib$.

$z$ can be represented by a polar representation as $r(\cos \theta+i\sin \theta)$,

when $r$ is the absolute value of $z$, $\sqrt{a^2 + b^2}$.

But how can I find $\theta$?

  • $\begingroup$ Consider the right triangle formed by the complex number in the Argand-Gauss plane and it's projections on the axis. $\endgroup$ – José Siqueira Nov 12 '13 at 17:21
  • $\begingroup$ In particular what is the definition of sine of theta in terms of the known sides of the above mentioned right triangle? $\endgroup$ – Adam Nov 12 '13 at 17:27

Consider the following Argand-diagram

enter image description here

The y-axis is the imaginary axis and the x-axis is the real one. The complex number in question is

$$x + yi$$

To figure out $\theta$, consider the right-triangle formed by the two-coordinates on the plane (illustrated in red). Let $\theta$ be the angle formed with the real axis.

$$\tan\theta = \frac{y}{x}$$

$$\implies \boxed{\tan^{-1}\left(\frac{y}{x}\right)}$$

The hypotenuse of the triangle will be

$$\sqrt{x^2 + y^2}$$


$$\sin\theta = \frac{y}{\sqrt{x^2 + y^2}}$$


So long as $a,b$ not both $0,$ the system of equations $$\begin{cases}\cos\theta=\frac{a}{\sqrt{a^2+b^2}}\\\sin\theta=\frac{b}{\sqrt{a^2+b^2}}\end{cases}$$ has a unique solution in the interval $[0,2\pi).$ In particular: if $a=0,$ then we have either $\theta=0$ or $\theta=\pi$ (determined by the second equation); if $b=0$, then we have either $\theta=\frac\pi2$ or $\theta=\frac{3\pi}2$ (determined by the first equation); if $a,b\ne 0,$ then either $\theta=\arctan\frac ba$ or $\theta=\pi+\arctan\frac ba$ (determined by the quadrant in which $a+ib$ lies).


I think, you can find it using inverse trigonometric functions: $$\varphi = arccos(\frac{a}{r})$$ $$\varphi = arcsin(\frac{b}{r})$$ where $$r = \sqrt{a^2+b^2}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.