# Thread: Two small things with Complex Numbers

1. ## Two small things with Complex Numbers

1)
I got:
$\displaystyle z=2+2i$ and I have to find the polar representation of z. Them find $\displaystyle z^4$ and $\displaystyle x+iy$ (when it's polar)...

My bid:
$\displaystyle w=rCos+i rSin$ to find the polar representation... right? So it would be something like: $\displaystyle w=r*cos(2)+r*sin(2i)$?

2a)
$\displaystyle w=x+iy$ find all values for w, when $\displaystyle w^2=-2i$

2b)
Find all z's when $\displaystyle z^2+(1+i)z+i=0$

2. Originally Posted by MissWonder
1)
I got:
$\displaystyle z=2+2i$ and I have to find the polar representation of z. Them find $\displaystyle z^4$ and $\displaystyle x+iy$ (when it's polar)...

My bid:
$\displaystyle w=rCos+i rSin$ to find the polar representation... right? So it would be something like: $\displaystyle w=r*cos(2)+r*sin(2i)$?

2a)
$\displaystyle w=x+iy$ find all values for w, when $\displaystyle w^2=-2i$

2b)
Find all z's when $\displaystyle z^2+(1+i)z+i=0$
$\displaystyle z = 2 + 2i$.

$\displaystyle r = \sqrt{2^2 + 2^2}$

$\displaystyle = \sqrt{8}$

$\displaystyle = 2\sqrt{2}$.

Since $\displaystyle z$ is in the first quadrant

$\displaystyle \theta = \arctan{\frac{2}{2}}$

$\displaystyle = \arctan{1}$

$\displaystyle = \frac{\pi}{4}$.

Thus

$\displaystyle z = 2 + 2i = 2\sqrt{2}\,\textrm{cis}\,\frac{\pi}{4}$.

To find $\displaystyle z^4$, use DeMoivre's Theorem.

$\displaystyle z^4 = (2\sqrt{2})^4\,\textrm{cis}\,\frac{4\pi}{4}$

$\displaystyle = 64\,\textrm{cis}\,{\pi}$

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

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

$\displaystyle = -64$.

3. ## Coool

I did not know how to do that :S Strange! our teatcher did it in another way.. But your explaination made sense

4. ## hint

I maybe understand 2a and 2b now.. $\displaystyle w^2*-2i=x+iy$ ??

5. Originally Posted by MissWonder
I maybe understand 2a and 2b now.. $\displaystyle w^2*-2i=x+iy$ ??
No, it would actually be

$\displaystyle w^2 = -2i$

$\displaystyle w^2 + 2i = 0$.

You could solve this using the quadratic formula, but the solution would involve $\displaystyle \sqrt{i}$.

Better to convert to polars and use DeMoivre's Theorem.

$\displaystyle w^2 = 2i = 2\,\textrm{cis}\,\frac{\pi}{2}$.

Supposing we knew what $\displaystyle w$ was, if we squared it we would have

$\displaystyle w^2 = r^2\,\textrm{cis}\,(2\theta)$.

Try to solve

$\displaystyle r^2\,\textrm{cis}\,(2\theta) = 2\,\textrm{cis}\,\frac{\pi}{2}$.

6. Try to solve

.

r=pi+cis ??

7. Originally Posted by MissWonder
Try to solve

.

r=pi+cis ??
Wouldn't it make sense that $\displaystyle r^2 = 2$ and $\displaystyle 2\theta = \frac{\pi}{2}$?

Solve for $\displaystyle r$ and $\displaystyle \theta$.

8. By the way,

$\displaystyle z = r\,\textrm{cis}\,\theta$

is shorthand for

$\displaystyle z = r(\cos{\theta} + i\sin{\theta})$.

9. r= 1.41421
and the other... hmm pi?

10. Hmm.. maybe rather cis=2/pi