# Thread: Two small things with Complex Numbers

1. ## Two small things with Complex Numbers

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

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

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

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

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

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

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

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

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

$= \sqrt{8}$

$= 2\sqrt{2}$.

Since $z$ is in the first quadrant

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

$= \arctan{1}$

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

Thus

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

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

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

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

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

$= 64(-1 + 0i)$

$= -64$.

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.. $w^2*-2i=x+iy$ ??

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

$w^2 = -2i$

$w^2 + 2i = 0$.

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

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

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

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

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

Try to solve

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

6. Try to solve

.

r=pi+cis ??

7. Originally Posted by MissWonder
Try to solve

.

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

Solve for $r$ and $\theta$.

8. By the way,

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

is shorthand for

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

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

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