Thread: Complex Numbers raised to the power?

If v = z(1)^6* z(2)^−4
determine |v| and angle(v). Also, obtain v in cartesian form.

where z1= 3 +i
z2 = -5 + 5i

I don't know how to raise a complex power to a power. I have searched all over but nothing online makes sense.
Please help me. Thanks!

So far I have done |(√10)^6 * (5√2)^-4| = .4

If that is correct please explain how to solve for angles.

Hey kmalik001.

If you have a complex number z = r*(cos(theta) + i*sin(theta)) then z^n = r^n * (cos(n*theta) + i*sin(n*theta)).

Originally Posted by kmalik001

If v = z(1)^6* z(2)^−4
determine |v| and angle(v). Also, obtain v in cartesian form.
where z1= 3 +i
z2 = -5 + 5i
If that is correct please explain how to solve for angles.

Frankly, I do not know what that says.

Is it

It is foolish to try to guess as to what you really meant to post.
So please answer this.

Okay well for the first z1^6 I got 1000cos(110.58) + isin(-270)

So 2 questions:

Should the angles be in radians?

How am i going to multiply this answer to when I get z2^-4. Because I can not change it to re^jtheta form because the angles in cos and sin are different

like one angle is 110.58 and the other is -270

Yes that is what is says Plato

Originally Posted by kmalik001

like one angle is 110.58 and the other is -270

Your mistake is to use degrees.

Let .

Now

and .

Just one question are you sure you did your second angle -3pi correct?

Because (-3pi/4) ^-4 is equal to 3.16*pi which isn't equal to 3*pi

I just checked on the calculator but I want to double check with you

The second part of this question asks me to solve for z1^3000. I tried the same method but when I type in 3000 power in calculator it doesn't allow me is there another method I could use. Taylor's method or binomial theorem I always get confused solving

Originally Posted by kmalik001

The second part of this question asks me to solve for z1^3000. I tried the same method but when I type in 3000 power in calculator it doesn't allow me is there another method I could use. Taylor's method or binomial theorem I always get confused solving

Are you very sure that is not given as

That is the only way makes sense as a question.

This is the exact question.

If w = z1^(3000) determine angle of w in the range [0, 2pi). Your answer should be correct to four decimal
places.

Originally Posted by kmalik001

This is the exact question.
If w = z1^(3000) determine angle of w in the range [0, 2pi). Your answer should be correct to four decimal places.

That was NOT what I asked you could it be that
Not but .

Yes, z1 = 3 + i is the question

Originally Posted by kmalik001

This is the exact question.

If w = z1^(3000) determine angle of w in the range [0, 2pi). Your answer should be correct to four decimal places.

Well then,

I understand the 3000* arctan(1/3)

but what did you do after that to get 3.92 like what does the ,2pi mean?