# How do I find z using the theorem of De Moivre

• Oct 1st 2011, 06:45 AM
terrorsquid
How do I find z using the theorem of De Moivre
I was going through some old work sheets and couldn't seem to do this one for some reason...

If $\displaystyle z^2=2-2i$ find $\displaystyle z$ using the theorem of De Moivre.

What I did:

$\displaystyle z=\sqrt{2-2i}$

$\displaystyle 8^{\frac{1}{4}} (cis\left(\frac{7\pi}{8}\right))$

I assume I would find values for $\displaystyle cos\left(\frac{7\pi}{8}\right)$ and $\displaystyle sin\left(\frac{7\pi}{8}\right)$ and then I would be done right? Assuming this is what I am supposed to do, how do I solve for the angle $\displaystyle \frac{7\pi}{8}$?

Thanks.
• Oct 1st 2011, 06:57 AM
Plato
Re: How do I find z using the theorem of De Moivre
Quote:

Originally Posted by terrorsquid
I was going through some old work sheets and couldn't seem to do this one for some reason...
If $\displaystyle z^2=2-2i$ find $\displaystyle z$ using the theorem of De Moivre.

Let $\displaystyle z^2=2-2i=\sqrt{8}\exp \left( {\frac{{ - \pi i}}{4}} \right)$
so $\displaystyle z=\omega = \sqrt[4]{8}\exp \left( {\frac{{ - \pi i}}{8}} \right)$

If $\displaystyle \varsigma = \exp \left( {\pi i} \right)$ then the two roots are $\displaystyle \omega\cdot\varsigma^k,~k=0,1~.$
• Oct 1st 2011, 07:02 AM
Prove It
Re: How do I find z using the theorem of De Moivre
Quote:

Originally Posted by terrorsquid
I was going through some old work sheets and couldn't seem to do this one for some reason...

If $\displaystyle z^2=2-2i$ find $\displaystyle z$ using the theorem of De Moivre.

What I did:

$\displaystyle z=\sqrt{2-2i}$

$\displaystyle 8^{\frac{1}{4}} (cis\left(\frac{7\pi}{8}\right))$

I assume I would find values for $\displaystyle cos\left(\frac{7\pi}{8}\right)$ and $\displaystyle sin\left(\frac{7\pi}{8}\right)$ and then I would be done right? Assuming this is what I am supposed to do, how do I solve for the angle $\displaystyle \frac{7\pi}{8}$?

Thanks.

$\displaystyle \displaystyle |2 - 2i| = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$ and $\displaystyle \displaystyle \arg{(2-2i)} = -\arctan{\frac{2}{2}} = -\frac{\pi}{4}$.

So $\displaystyle \displaystyle 2 - 2i = 2\sqrt{2}\,\textrm{cis}\,\left(-\frac{\pi}{4}\right)$.

You have

\displaystyle \displaystyle \begin{align*} z^2 &= 2\sqrt{2}\,\textrm{cis}\left(-\frac{\pi}{4}\right) \\ z &= \left[2\sqrt{2}\,\textrm{cis}\left(-\frac{\pi}{4}\right)\right]^{\frac{1}{2}} \\ &= \sqrt{2\sqrt{2}}\,\textrm{cis}\left[\frac{1}{2}\left(-\frac{\pi}{4}\right)\right] \\ &= \sqrt[4]{8}\,\textrm{cis}\left(-\frac{\pi}{8}\right) \\ &= \sqrt[4]{8}\left[\cos{\left(-\frac{\pi}{8}\right)} + i\sin{\left(-\frac{\pi}{8}\right)}\right] \\ &= \sqrt[4]{8}\left[\cos{\left(\frac{\pi}{8}\right)} - i\sin{\left(\frac{\pi}{8}\right)}\right]\end{align*}

You will need to use half angle identities to evaluate these exactly.

There is also another square root. You should know that square roots are evenly spaced about a circle, so these two square roots will have the same magnitude and are separated by an angle of $\displaystyle \displaystyle \pi$, so the other square root is

\displaystyle \displaystyle \begin{align*} \sqrt[4]{8}\,\textrm{cis}\left(\pi - \frac{\pi}{8}\right) &= \sqrt[4]{8}\left[\cos{\left(\pi - \frac{\pi}{8}\right)} + i\sin{\left(\pi - \frac{\pi}{8}\right)}\right] \\ &= \sqrt[4]{8} \left[ -\cos{ \left(\frac{\pi}{8}\right) } + i \sin{ \left ( \frac{\pi}{8}\right) } \right]\end{align*}

and you should have those values from the first square root :)