# Complex numbers and modulus

1. Nov 15, 2005

### Atilla1982

I have z=-(1/2)-(sqrt3/2)i
r=|z|

is this right?

r=cos*2Pi/3+i*sin*Pi/3 = 1 + sqrt3/2*i

Now I have to find Z^2004, how do I do that?

2. Nov 15, 2005

### dextercioby

Nope. The modulus of a complex number has to be real.

Daniel.

3. Nov 15, 2005

### dextercioby

Show that the trigonometric form of your "z" is

$$z=\cos 210\mbox{deg} \ +i\sin 210\mbox{deg}$$

Daniel.

4. Nov 15, 2005

### benorin

no

No. If $$z=a+bi$$ where a and b are real, then $$\left| z \right| = \sqrt{a^2+b^2} \geq 0$$ for any complex number z. Notably, $$\left| z \right|$$ is always a positive real number, and hence your answer for r cannot be correct.

If $$z = -\frac{1}{2}-\sqrt{\frac{3}{2}}i$$, then $$r=\left| z \right| = \sqrt{\left( -\frac{1}{2} \right) ^2 + \left( -\sqrt{\frac{3}{2}} \right) ^2} = \sqrt{ \frac{1}{4}+ \frac{3}{2}} =\frac{1}{2}\sqrt{7}$$

5. Nov 15, 2005

### HallsofIvy

Staff Emeritus
$$r= \sqrt{\left(\frac{1}{2}\right)^2+ \left(\frac{\sqrt{3}}{2}\right)^2}$$
which, as Dextercioby and denorin point out, is a real number.
$$tan(\theta)= \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}}= \sqrt{3}$$
Since both real and imaginary parts of negative (the negatives disappear in the fraction) the angle is in the 3rd quadrant.
(Thanks, Dextercioby. One of these days, I really need to learn to count!)

Once you know r and $\theta$,
$$(r(cos(\theta)+ i sin(\theta))^n= r^n(cos(n\theta)+ i sin(n\theta))$$

Last edited: Nov 15, 2005
6. Nov 15, 2005

### dextercioby

You mean the third quadrant, right...?

Daniel.