Calculating Powers of Complex Numbers in the Third Quadrant

[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?

 

[dextercioby]

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

Daniel.

 

[dextercioby]

Show that the trigonometric form of your "z" is

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

Daniel.

 

[benorin]

no

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

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}

 

[HallsofIvy]

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?

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))

 

[dextercioby]

You mean the third quadrant, right...?

Daniel.