# Help with complex numbers in polar and cartesian form

• Dec 9th 2010, 05:52 AM
mariasmiles25
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.
• Dec 9th 2010, 06:11 AM
TheEmptySet
Quote:

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 $z=x+iy$ then

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

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

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

Then just use the laws of exponents

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

So

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

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

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

$\displaystyle e^{i\theta}=\cos(\theta)+i\sin(\theta)$
• Dec 9th 2010, 06:14 AM
Plato
Let me be candid. I am in the camp of those who do not like the notation $\sqrt{1-\sqrt3~i}$, i.e. using the radical sign in this way.
We prefer asking for the square roots of ${1-\sqrt3~i}$.
$1-\sqrt3~i =2\exp\left(\frac{-\pi i}{3}\right)$ so the first square root is $\sqrt2\exp\left(\frac{-\pi i}{6}\right)$.
We know that $\frac{2\pi}{2}=\pi$ 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.