# Converting to polar form

Write each of the given numbers in the polar form $re^{i\theta}$.

a.) $\frac{1-i}{3}$

b.) $-8\pi (1+\sqrt 3 i)$

For a, I got:

r = $\frac{\sqrt 2}{3}$ and $e^{i7\pi /2}$ since $\frac{\frac{1}{3}}{\frac{\sqrt 2}{3}} = \frac{\sqrt 2}{2}$ and $\frac{-\frac{1}{3}}{\frac{\sqrt 2}{3}} = -\frac{\sqrt 2}{2}$, but the answer for this is $\frac{\sqrt 2}{3}e^{-i\frac{\pi}{4}}$, how did they get that?

For b, I got: $-16\pi e^{i\frac{\pi}{3}}$ but the answer for this is $16\pi e^{-i\frac{2\pi}{3}}$.

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## 1 Answer

Following two points need to be used here:

(1) The principal value of $\arctan \frac yx$ lies in $[-\frac\pi2,\frac\pi2]$

(2)From this, we need to identify the principal value of the argument of a complex number.

Let $$\frac{1-i}3=re^{i\theta}=r(\cos\theta+ i\sin\theta)$$ where $r>0$

Equating the real & the imaginary parts,

$r\cos\theta=\frac13$ and $r\sin\theta=-\frac13$

Squaring and adding we get, $r^2=(\frac13)^2+(-\frac13)^2=\frac29$

So, $r=\frac{\sqrt2}3$ and $\cos\theta=\frac{\frac13}{\frac{\sqrt2}3}=\frac1{\sqrt2}$ and $\sin\theta=\frac{-\frac13}{\frac{\sqrt2}3}=-\frac1{\sqrt2}$

so $\tan\theta=-1$ and $\theta$ lies in the 4th quadrant hence $\theta=-\frac\pi4$

Similarly, if we put $-8\pi(1+\sqrt3i)=r(\cos\theta+ i\sin\theta)$ where $r>0$

Applying the same method, $r^2=(-8\pi)^2+(-8\pi\sqrt3)^2=(8\pi)^24\implies r=16\pi$

So, $\cos\theta=\frac{-8\pi}{16\pi}=-\frac12$ and $\sin\theta=\frac{-8\pi\sqrt3}{16\pi}=-\frac{\sqrt3}2$

so, $\tan\theta=\sqrt3$ and $\theta$ lies in the 3th quadrant hence $\theta=-\pi+\frac\pi3=-\frac{2\pi}3$

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But if you take the square root of $\frac{2}{9}$ you will get $r=\frac{\sqrt2}{3}$ – Q.matin Jan 20 '13 at 7:13
@Q.matin, rectified the type error. – lab bhattacharjee Jan 20 '13 at 7:15
Also, if it were in the fourth quadrant why arent we using $7 \pi /4$ instead of $-\pi /4$? – Q.matin Jan 20 '13 at 7:15
@Q.matin, if you meant $\frac{7\pi}4,$ please find en.wikipedia.org/wiki/Complex_number#Polar_form, the values of the argument for the different signs of $x,y$ – lab bhattacharjee Jan 20 '13 at 7:19
@Q.matin, my pleasure. – lab bhattacharjee Jan 20 '13 at 7:29