# Thread: Polar form of complex numbers

1. ## Polar form of complex numbers

Hi

here

2. $\displaystyle -4.88 - 3.17i$.

Notice that this is in the third quadrant.

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

$\displaystyle = \sqrt{23.8144 + 10.0489}$

$\displaystyle = \sqrt{33.8633}$

$\displaystyle \approx 5.81922$.

The angle is in the third quadrant.

So it's calculated as

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

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

Therefore:

$\displaystyle -4.88 - 3.17i \approx 5.81922 \,\textrm{cis}\,(-146.993^{\circ})$.

3. First you need to convert to polars.

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

$\displaystyle = \sqrt{1 + 3}$

$\displaystyle = \sqrt{4}$

$\displaystyle = 2$.

Notice that the angle is in the fourth quadrant.

So it's calculated by

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

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

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

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

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

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

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

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

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

$\displaystyle = 64(1 + 0i)$

$\displaystyle = 64$.

4. thanks but why you wrote ( cis ) no e

5. $\displaystyle \textrm{cis}\,\theta$ is just a shorthand way of writing $\displaystyle \cos{\theta} + i\sin{\theta}$.

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

So I could have written $\displaystyle e$, I just prefer using the $\displaystyle \textrm{cis}$ notation, as I find it easier to convert between polars and cartesians.

6. Hi prove it