0
$\begingroup$

Write an equation that relates $\arg(z)$ to $\arg(1/z)$, $z\not = 0$.

I don't have a clear idea of how attack this type of exercise.

I know if $z=r(\cos\theta +i\sin\theta$) or equivalent to $z=x+iy$ then the $\arg(z)=\theta$ and we can calculate $\theta$ using the fact $\tan\theta=\frac{y}{x}\implies\theta=\tan^{-1}(\frac{y}{x})$ But, here I'm stuck. Can someone help me?

$\endgroup$
1
$\begingroup$

Write $z$ in the polar form as $z=re^{i\theta}$. Then $1/z=\frac{1}{r}e^{-i\theta}$. So $\arg(1/z)=-\arg(z)$.

$\endgroup$
  • $\begingroup$ Even for $z=-1$? $\endgroup$ – Lord Shark the Unknown Oct 12 '18 at 15:18
  • $\begingroup$ Yes, because $\arg(1/-1)=\arg(-1)=\pi=-\pi=-\arg(-1)$. The angles $\pi$ and $-\pi$ are equivalent. Although you do bring up a good point, which is that you may need to add or subtract $2\pi$ to get the argument in the usual range $\endgroup$ – HackerBoss Oct 12 '18 at 15:49
  • $\begingroup$ Ah $\pi=-\pi$. You live and learn! $\endgroup$ – Lord Shark the Unknown Oct 12 '18 at 15:50
0
$\begingroup$

So if $$z = r(\cos \theta + i \sin \theta)$$ what is the expression for $1/z$? You can use the usual trick for division by complex numbers: multiple numerator and denominator by the complex conjugate.

Now once you have an expression for $1/z$, you should be able to infer the correct angle.

$\endgroup$

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.