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What is the polar form of $-4+4i$?

I am getting argument as negative $45^{\circ}$.

So is $-45^{\circ}$ my final argument?

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    $\begingroup$ Wrong direction for negative angles. $-45^\circ$ is in the fourth quadrant. Your point is in the second quadrant. $\endgroup$ – user4894 Aug 21 '16 at 7:10
  • $\begingroup$ The correct method is described here: math.stackexchange.com/a/256333 $\endgroup$ – David K Aug 21 '16 at 7:34
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The modulus gives you the radius so $$ r=\sqrt{16+16}=4\sqrt{2} $$ And your angle is given by $$ \arctan(-1)=-\pi/4 $$ Then you need to see what quadrant you are in, which in your case is negative real and positive imaginary components, so in the second quadrant, yielding an addition of $\pi$ and an argument of $3\pi/4$.

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Angle or argument is in second quadrant $( x <0, y>0)$. The angle is found from atan2 function which is sensitive to sign of x,y components, placing it in second quadrant. If not careful you may end up in the fourth quadrant choice.

$$ (+,+)\rightarrow Q1;\, (-,+)\rightarrow Q2; \, (-,-)\rightarrow Q3 ; \, Q(+,-)\rightarrow Q4\,$$

So in polar form

$$ r e^{i \theta} = 4 \sqrt 2 e^{i\,3 \pi/4 }. $$

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Let $z=x+yi$ , so $x=-4$ and $y=4$ .

Magnitude of $z$ is given by $r=|z|=\sqrt{x^2+y^2}=\sqrt{16+16}=4\sqrt{2}$

Since $x<0$ and $y\ge 0$ the argument of $z$ is given by $\varphi=\arctan(\frac{y}{x})+\pi =-\frac{\pi}{4}+\pi=\frac{3\pi}{4}$ .

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