Two small things with Complex Numbers

**MissWonder** · September 27th 2009, 06:24 AM

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 $z^2+(1+i)z+i=0$

**Prove It** · September 27th 2009, 06:41 AM

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.

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

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

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

$= 64(-1 + 0i)$

$= -64$ .

**MissWonder** · September 27th 2009, 07:41 AM

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

**MissWonder** · September 27th 2009, 08:22 AM

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

**Prove It** · September 27th 2009, 08:30 AM

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

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