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.

Re: Complex Numbers raised to the power?

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)).

Re: Complex Numbers raised to the power?

Quote:

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 $\displaystyle (3+i)^6\cdot(-5+5i)^{-4}~?$

It is foolish to try to guess as to what you really meant to post.

So please answer this.

Re: Complex Numbers raised to the power?

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

Re: Complex Numbers raised to the power?

like one angle is 110.58 and the other is -270

Re: Complex Numbers raised to the power?

Yes that is what is says Plato

Re: Complex Numbers raised to the power?

Quote:

Originally Posted by

**kmalik001** like one angle is 110.58 and the other is -270

Your mistake is to use degrees.

Let $\displaystyle \theta=\arctan\left(\frac{1}{3}\right)$.

Now $\displaystyle z_1=\sqrt{10}\exp(\theta i)~\&~z_2=5\sqrt{2}\exp\left(\frac{-3\pi i}{4}\right)$

$\displaystyle (z_1)^6=10^3\exp(6\theta i)$ and $\displaystyle (z_2)^{-4}=5^{-4}(2^{-2})\exp\left({-3\pi i}\right)$.

Re: Complex Numbers raised to the power?

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

Re: Complex Numbers raised to the power?

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

Re: Complex Numbers raised to the power?

Quote:

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 $\displaystyle z_1$ is not given as $\displaystyle \sqrt{3}+i~?$

That is the only way $\displaystyle (z_1)^{3000}$ makes sense as a question.

Re: Complex Numbers raised to the power?

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.

Re: Complex Numbers raised to the power?

Quote:

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 $\displaystyle z_1=\sqrt{3}+i~?$

Not $\displaystyle 3+i$ but $\displaystyle \sqrt{3}+i$.

Re: Complex Numbers raised to the power?

Yes, z1 = 3 + i is the question

Re: Complex Numbers raised to the power?

Quote:

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, $\displaystyle \text{arg}\left( {3 + i} \right)^{3000}=\bmod \left( {3000\arctan \left( {\frac{1}{3}} \right),2\pi } \right) = {\text{3}}{\text{.9243}} $

Re: Complex Numbers raised to the power?

I understand the 3000* arctan(1/3)

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