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