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Write $z=5-5i$ in the polar form with $z=Arg(z)$

My work

Let $z=5-5i\in \mathbb{C}$ then

$r=\sqrt{5^2+5^2}=5\sqrt{2}$

$\theta=tan^{-1}(\frac{-5}{5})=\frac{-\pi}{4}$

By definition, $-\pi<Arg(z)\leq \pi$ then

enter image description here

I have problem trying to find that angle. Can someone help me?

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    $\begingroup$ You already have the correct answer in $r=5\sqrt2$ and $\theta =-\frac {\pi}4$. What else do you need? $\endgroup$ – Mohammad Zuhair Khan Oct 12 '18 at 16:49
  • $\begingroup$ The angle when $\theta=Arg(z)$ @MohammadZuhairKhan $\endgroup$ – Bvss12 Oct 12 '18 at 16:52
  • $\begingroup$ I assume you want the angle that fits in $[0,2\pi]$ or atleast that is how I remember I was taught in high-school. Well, remember that $e^{ix}$ has a periodicity of $2\pi$ so you can just add $2\pi$ to $-\frac{\pi}{4}$ and get the angle you seek. Of course you can also look at your graph and see that you are in the fourth quadrant and well, it's at a distance of $2\pi$ from the right one. $\endgroup$ – Zacky Oct 12 '18 at 17:04
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Actually, you already finished your work. You found the the module is $5\sqrt{2}$, the argument is $-\frac{\pi}{4}$, so the number is $z=5\sqrt{2}e^{-i\frac{\pi}{4}}=5\sqrt{2}(\cos(-\frac{\pi}{4})+i\sin(-\frac{\pi}{4}))=5\sqrt{2}(\cos(\frac{\pi}{4})-i\sin(\frac{\pi}{4}))$.

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  • $\begingroup$ I need the other angle, when $\theta= Arg(z)$, the angle $\theta=\frac{-\pi}{4}$ is when $Arg(z)\not = \theta$ $\endgroup$ – Bvss12 Oct 12 '18 at 16:54
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    $\begingroup$ Didn't you write you need the angle between $-\pi$ and $\pi$? Well, $-\frac{\pi}{4}$ is exactly there. I don't understand the problem. $\endgroup$ – Mark Oct 12 '18 at 16:56
  • $\begingroup$ I need the principal argument of $z$... $\endgroup$ – Bvss12 Oct 12 '18 at 16:58
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    $\begingroup$ Yes, the principal argument is the argument in the interval $(-\pi,\pi]$. And indeed we have $-\frac{\pi}{4}\in(-\pi,\pi]$. This is the principal argument. $\endgroup$ – Mark Oct 12 '18 at 16:59
  • $\begingroup$ Okay, thanks... is a little confused. Thanks $\endgroup$ – Bvss12 Oct 12 '18 at 17:01

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