0
$\begingroup$

Considering two complex numbers $z_1$ and $z_2$ in the form $z=r(\cos(\theta)+i\sin(\theta))$

There is this formula $z_1 z_2=r_1 r_2 (\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2))$

So this relation should hold : $\arg(z_1 z_2)=\arg(z_1)+\arg(z_2)$

But if I consider $z_1=-1=\cos(\pi)+i\sin(\pi)$ and $z_2=i=\cos(\frac{\pi}{2})+i\sin (\frac{\pi}{2})$

$z_1 z_2=-i=\cos(-\frac{\pi}{2})+i\sin(-\frac{\pi}{2})$

While $\arg(z_1)+\arg(z_2)= \pi+ \frac {\pi}{2}=\frac{3 \pi}{2}\neq \frac{-\pi}{2}$

The angle is actually the same but I get two different results. Is this normal or am I missing something?

Thanks for your help

$\endgroup$
  • $\begingroup$ Remeber that arguments are only defined up to $\pm k 2\pi$ where $k$ is a natural number. In this example, starting at the $x$-axis and going anti-clockwise $3\pi/2$ is the same as going clockwise $\pi/2$ (the $-$ sign reverses the orientation). $\endgroup$ – James Dec 27 '15 at 15:05
  • $\begingroup$ $$\frac{3\pi}2=-\frac\pi2\mod2\pi$$ $\endgroup$ – Oskar Limka Dec 27 '15 at 15:07
2
$\begingroup$

Yes, this is normal. Note that $\sin$ and $\cos$ are periodic with period $2\pi$ thus a difference of $2\pi$ or an integral multiple of $2 \pi$ has not real effect.

There are several ways out of this.

  • One can consider the argument not as one real number but as the set of all possible arguments. The set of arguments of $z_1z_2$ is the set of sums of arguments of $z_1$ and arguments of $z_2$.

  • One can fix the argument to lie in a fixed interval of length $2\pi$, for example in $(-\pi,\pi]$. Then one calls this the principal argument. It is then not always true that the sum of the principal arguments is the principal argument of the product. One might have to account for $2 \pi$ to get the sum back into the "good" interval.

$\endgroup$
0
$\begingroup$

$\arg(z_1z_2)=\arg(z_1)+\arg(z_2)+2k\pi$ where $k\in\mathbb{Z}$ i.e., $\arg(z_1z_2)=(\arg(z_1)+\arg(z_2))\mod 2\pi.$

$\endgroup$

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.