Complex number cartesian form

Hello everyone,

Need a bit of help.

(-2-2i)^5

I want to turn this into Cartesian form using the modulus.

Into the form
z^n=r^n(cos n0 +i sin n0)

So far I have r=2sqrt(2)

The angle ive worked out as pi/4 or 45 degrees (im not 100% confident here).

Now im a bit unsure how I put it together.

Thanks for any help!

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asked by mrbob
  1. yes r^2 = 4 + 4 = 8
    so
    r = sqrt 8 = 2 sqrt 2

    - 2 - 2 i is in quadrant 3, 45 degrees below -x axis

    Theta = pi + pi/4 = 5 pi/4
    or 180 + 45 = 225 degress
    so

    x = 8^.5 e^(5 pi i/4)

    now raise to power five
    raise (8^.5)^5 = 8^2.5
    multiply angle by 5
    25 pi/4 = 24 pi/4 + pi/4
    which is
    6 pi + pi/4
    which is just pi/4 because every 2 pi is all the way around
    so
    8^2.5 (cos 45 + i sin 45

    8^2.5 (sqrt2 /2 + i sqrt 2/2)
    2^7.5 (2^.5 + i 2^.5)/2
    (2^8 + i 2^8)/2
    2^7 + i 2^7
    or
    128 + 128 i

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    posted by Damon
  2. Well, that was entertaining :)

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    posted by Damon
  3. Thanks Damon, really helped.

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    posted by mrbob

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