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As it is said in the title.

The number is $4-4i$

I did this

$4-4i=\sqrt 2\,(1+i)=\cos(\pi/4)+i \sin(\pi/4)=\sqrt{2}e^{i(\pi/4)}$

$|4-4i|=\sqrt{2}$, $arg(4-4i)=(\pi/4)+2\pi$

but it's wrong

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  • $\begingroup$ In the FAQ section there are directions for writing mathematics with LaTeX in this site. $\endgroup$
    – DonAntonio
    Mar 29 '13 at 14:47
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At first we calculate the $r$, as $4^2 +4^2 = 32$ $r$ will be $\sqrt{32}$. Because $$4-4i=4(1-i)=r \cdot \exp(i \varphi)$$ when making a sketch we see that the angle is $$\frac{7\pi}{4}$$ so the whole number will be $$4-4i= \sqrt{32} \cdot \exp\left(i \cdot \frac{7\pi}{4}\right)$$

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$$4+4i=4(1-i)=4\sqrt 2\left(\cos\frac{\pi}{4}-i\sin\frac{\pi}{4}\right)=4\sqrt 2e^{-i\pi/4}$$

And, of course, you should add multiples of $\,2\pi\,$ to the argument, unless you want one single value.

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Let $4-4i=r(\cos\theta+i\sin\theta)$ where $r\ge 0$

Equating the real & the imaginary parts

$4=r\cos\theta, -4=r\sin\theta$

Squaring and adding we get $r^2=4^2+4^2=32\implies r=4\sqrt2$

$\implies \cos\theta=\frac1{\sqrt2}>0, \sin\theta=-\frac1{\sqrt2}<0 $

From the definition of $\arctan\frac yx,\theta= \arctan\frac{(-1)}1=\arctan(-1)=-\frac{\pi}4$ as the principal value of arctan lies $\in\left[-\frac\pi2, \frac\pi2\right]$

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