# Thread: Help with complex numbers in polar and cartesian form

1. ## Help with complex numbers in polar and cartesian form

How do you write the following equations in polar form z=re^io and find the values. Then write the answer in cartesian form x+iy

√(1-i√3)

cube√(-4(√2)+4√(2i))

*Sorry, I posted this in the wrong thread earlier.

2. Originally Posted by mariasmiles25
How do you write the following equations in polar form z=re^io and find the values. Then write the answer in cartesian form x+iy

√(1-i√3)

cube√(-4(√2)+4√(2i))

*Sorry, I posted this in the wrong thread earlier.
If $\displaystyle z=x+iy$ then

$\displaystyle \displaystyle z=\sqrt{x^2+y^2}e^{i\tan^{-1}\left( \frac{y}{x}\right)}$

Remember that $\displaystyle \displaystyle \tan^{-1}\left( \frac{y}{x}\right)$

Only gives angles in the range $\displaystyle \left(\frac{\pi}{2},\frac{\pi}{2} \right)$ if $\displaystyle x < 0$ you need to $\displaystyle \pi$ to the angle gives

Then just use the laws of exponents

$\displaystyle \displaystyle z=1-i\sqrt{3}=2e^{-i\frac{\pi}{3}}$

So

$\displaystyle \displaystyle (z)^{\frac{1}{2}}=(1-i\sqrt{3})^{\frac{1}{2}}=(2e^{-i\frac{\pi}{3}})^{\frac{1}{2}}=2^{\frac{1}{2}}e^{-i\frac{\pi}{6}+ki\pi}$

Where $\displaystyle k=0,1$ notice that when $\displaystyle k=2$ you get the same values as when $\displaystyle k=0$

This will give you two answers Now use Euler's formula to convert back.

$\displaystyle \displaystyle e^{i\theta}=\cos(\theta)+i\sin(\theta)$

3. Let me be candid. I am in the camp of those who do not like the notation $\displaystyle \sqrt{1-\sqrt3~i}$, i.e. using the radical sign in this way.
We prefer asking for the square roots of $\displaystyle {1-\sqrt3~i}$.
$\displaystyle 1-\sqrt3~i =2\exp\left(\frac{-\pi i}{3}\right)$ so the first square root is $\displaystyle \sqrt2\exp\left(\frac{-\pi i}{6}\right)$.
We know that $\displaystyle \frac{2\pi}{2}=\pi$ so add that to the argument of the first root to get the second square root.

To get the cube roots divide by three instead of two. There are three roots.