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

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 then

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

• You already have the correct answer in $r=5\sqrt2$ and $\theta =-\frac {\pi}4$. What else do you need? – Mohammad Zuhair Khan Oct 12 '18 at 16:49
• The angle when $\theta=Arg(z)$ @MohammadZuhairKhan – Bvss12 Oct 12 '18 at 16:52
• 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. – Zacky Oct 12 '18 at 17:04

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 $$z=5\sqrt{2}e^{-i\frac{\pi}{4}}=5\sqrt{2}(\cos(-\frac{\pi}{4})+i\sin(-\frac{\pi}{4}))=5\sqrt{2}(\cos(\frac{\pi}{4})-i\sin(\frac{\pi}{4}))$$.
• I need the other angle, when $\theta= Arg(z)$, the angle $\theta=\frac{-\pi}{4}$ is when $Arg(z)\not = \theta$ – Bvss12 Oct 12 '18 at 16:54
• 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. – Mark Oct 12 '18 at 16:56
• I need the principal argument of $z$... – Bvss12 Oct 12 '18 at 16:58
• 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. – Mark Oct 12 '18 at 16:59