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Math Help - Complex Numbers problem

  1. #1
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    Cool Complex Numbers problem

    Calculate: (1+3i)^9

    How do I solve this? I have tried without any success. Thanks!
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  2. #2
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    Quote Originally Posted by sf1903 View Post
    Calculate: (1+3i)^9
    1 + i\sqrt 3  = 2\cos \left( {\frac{\pi }<br />
{3}} \right) + i2\sin \left( {\frac{\pi }<br />
{3}} \right)
    2^9=512 and 9\left(\frac{\pi}{3}\right)=3\pi
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    Hint : 2e^{i\pi/3}=1+\sqrt{3}i.
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  4. #4
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    Quote Originally Posted by Plato View Post
    1 + i\sqrt 3 = 2\cos \left( {\frac{\pi }<br />
{3}} \right) + i2\sin \left( {\frac{\pi }<br />
{3}} \right)
    2^9=512 and 9\left(\frac{\pi}{3}\right)=3\pi
    If im correct the answer should be -512

    512*Cos(3π)+i*512*Sin(3π) = -512 + 0 = -512

    Or am i out in the blue?
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  5. #5
    MHF Contributor Bruno J.'s Avatar
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    Yes! (2e^{i\pi/3})^9=512e^{3i\pi}=-512.
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