# Complex Numbers raised to the power?

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• Feb 12th 2013, 03:48 PM
kmalik001
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.

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

If that is correct please explain how to solve for angles.
• Feb 12th 2013, 04:09 PM
chiro
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)).
• Feb 12th 2013, 04:16 PM
Plato
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.
• Feb 12th 2013, 04:20 PM
kmalik001
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
• Feb 12th 2013, 04:21 PM
kmalik001
Re: Complex Numbers raised to the power?
like one angle is 110.58 and the other is -270
• Feb 12th 2013, 04:21 PM
kmalik001
Re: Complex Numbers raised to the power?
Yes that is what is says Plato
• Feb 12th 2013, 04:56 PM
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 $\displaystyle (z_2)^{-4}=5^{-4}(2^{-2})\exp\left({-3\pi i}\right)$.
• Feb 12th 2013, 05:07 PM
kmalik001
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
• Feb 12th 2013, 05:25 PM
kmalik001
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
• Feb 12th 2013, 05:34 PM
Plato
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.
• Feb 12th 2013, 05:49 PM
kmalik001
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.
• Feb 12th 2013, 06:00 PM
Plato
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 $\displaystyle \sqrt{3}+i$.
• Feb 12th 2013, 06:04 PM
kmalik001
Re: Complex Numbers raised to the power?
Yes, z1 = 3 + i is the question
• Feb 12th 2013, 06:22 PM
Plato
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, $\displaystyle \text{arg}\left( {3 + i} \right)^{3000}=\bmod \left( {3000\arctan \left( {\frac{1}{3}} \right),2\pi } \right) = {\text{3}}{\text{.9243}}$
• Feb 13th 2013, 04:06 AM
kmalik001
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?
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