# Polar form of complex numbers

• Apr 8th 2010, 06:06 AM
r-soy
Polar form of complex numbers
• Apr 8th 2010, 06:18 AM
Prove It
$-4.88 - 3.17i$.

Notice that this is in the third quadrant.

Its length is $\sqrt{(-4.88)^2 + (-3.17)^2}$

$= \sqrt{23.8144 + 10.0489}$

$= \sqrt{33.8633}$

$\approx 5.81922$.

The angle is in the third quadrant.

So it's calculated as

$\theta = -180^{\circ} + \arctan{\frac{3.17}{4.88}}$

$\theta \approx -146.993^{\circ}$.

Therefore:

$-4.88 - 3.17i \approx 5.81922 \,\textrm{cis}\,(-146.993^{\circ})$.
• Apr 8th 2010, 06:25 AM
Prove It
First you need to convert to polars.

$|1 - i\sqrt{3}| = \sqrt{1^2 + (-\sqrt{3})^2}$

$= \sqrt{1 + 3}$

$= \sqrt{4}$

$= 2$.

Notice that the angle is in the fourth quadrant.

So it's calculated by

$\theta = 2\pi - \arctan{\frac{\sqrt{3}}{1}}$

$= 2\pi - \arctan{\sqrt{3}}$

$= 2\pi - \frac{\pi}{3}$

$= \frac{5\pi}{3}$.

So $1 - i\sqrt{3} = 2\,\textrm{cis}\,\left(\frac{5\pi}{3}\right)$

Therefore $(1 - i\sqrt{3})^6 = \left[2\,\textrm{cis}\,\left(\frac{5\pi}{3}\right)\right]^6$

$= 2^6\,\textrm{cis}\,\left(\frac{6\cdot 5\pi}{3}\right)$

$= 64 \,\textrm{cis}\,\left(10\pi\right)$

$= 64(\cos{10\pi} + i\sin{10\pi})$

$= 64(1 + 0i)$

$= 64$.
• Apr 8th 2010, 10:53 AM
r-soy
thanks but why you wrote ( cis ) no e
• Apr 8th 2010, 09:22 PM
Prove It
$\textrm{cis}\,\theta$ is just a shorthand way of writing $\cos{\theta} + i\sin{\theta}$.

As we know, $\cos{\theta} + i\sin{\theta} = e^{i\theta}$.

So I could have written $e$, I just prefer using the $\textrm{cis}$ notation, as I find it easier to convert between polars and cartesians.
• Apr 9th 2010, 06:21 AM
r-soy
Hi prove it