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Math Help - 3 Complex Number questions

  1. #1
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    3 Complex Number questions

    1. Find (2+2i)^8 in Cartesian form
    2. Square roots of -i
    3. Show the following graphically |z+1|=|z-1| (abs value)

    I am lost on 1 but my answer for 2 is i, sqrt[2](1-i)/2
    as for 3, is it just a unit circle?

    Thanks!
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  2. #2
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    1) (2+2i)^8 = ((2+2i)^2)^4 = (i8)^4=4096.

    2) Suppose (a+ib)^2 = a^2-b^2 + 2iab = -i. Equating real and imaginary parts gives |a|=|b| and ab=-1/2. The (two!) solutions are then {\sqrt{2}\over2}(1-i) and {\sqrt{2}\over2}(i-1).

    3) This is the set of points z whose distance from 1 is the same as their distance from -1. The locus of points equidistant from two points is a straight line perpendicular to the line joining the two points and passing through its midpoint. In this case, it is the line on the imaginary axis.
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  3. #3
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    for 1), the problem says to find it in the form x+iy with x and y real.. 4096 is the answer straight up?
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  4. #4
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    Yes, with x=4096 and y=0 :]
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  5. #5
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    ahh obviously!! thanks!

    i hate these complex roots and stuff... too confusing
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  6. #6
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    wait the way i tried to do b didnt work.. where did i go wrong?

    -i in polar coordinates is: e^(i3pi/2)

    so -1= cos(3pi/2) + isin (3pi/2)
    and the the other root is = cos(5pi/2) + isin (5pi/2)


    so then you divide each angle by 2 right? and then solve?
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  7. #7
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    hmm.. one more problem i can't figure out

    Let z= e^(2+ipi/4)

    what is the conjugate of z? (I htink its called a conjugate.. where there is a dash on top of the z?)
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  8. #8
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    It is indeed called conjugate. Observe that \bar{z} := x-iy and e^{i\theta} = \cos \theta + i \sin \theta so the conjugate of e^{i\theta} is \cos\theta - i \sin\theta=\cos\theta+i\sin(-\theta) = e^{-i\theta}. So the conjugate of your z is e^{2-ipi/4}.
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