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Math Help - Help with complex numbers in polar and cartesian form

  1. #1
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    Help with complex numbers in polar and cartesian form

    How do you write the following equations in polar form z=re^io and find the values. Then write the answer in cartesian form x+iy

    √(1-i√3)

    cube√(-4(√2)+4√(2i))


    *Sorry, I posted this in the wrong thread earlier.
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  2. #2
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    Quote Originally Posted by mariasmiles25 View Post
    How do you write the following equations in polar form z=re^io and find the values. Then write the answer in cartesian form x+iy

    √(1-i√3)

    cube√(-4(√2)+4√(2i))


    *Sorry, I posted this in the wrong thread earlier.
    If z=x+iy then

    \displaystyle z=\sqrt{x^2+y^2}e^{i\tan^{-1}\left( \frac{y}{x}\right)}

    Remember that \displaystyle \tan^{-1}\left( \frac{y}{x}\right)

    Only gives angles in the range \left(\frac{\pi}{2},\frac{\pi}{2} \right) if x < 0 you need to \pi to the angle gives

    Then just use the laws of exponents

    \displaystyle z=1-i\sqrt{3}=2e^{-i\frac{\pi}{3}}

    So

    \displaystyle (z)^{\frac{1}{2}}=(1-i\sqrt{3})^{\frac{1}{2}}=(2e^{-i\frac{\pi}{3}})^{\frac{1}{2}}=2^{\frac{1}{2}}e^{-i\frac{\pi}{6}+ki\pi}

    Where k=0,1 notice that when k=2 you get the same values as when k=0

    This will give you two answers Now use Euler's formula to convert back.

    \displaystyle e^{i\theta}=\cos(\theta)+i\sin(\theta)
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    Let me be candid. I am in the camp of those who do not like the notation \sqrt{1-\sqrt3~i}, i.e. using the radical sign in this way.
    We prefer asking for the square roots of {1-\sqrt3~i}.
     1-\sqrt3~i =2\exp\left(\frac{-\pi i}{3}\right) so the first square root is  \sqrt2\exp\left(\frac{-\pi i}{6}\right).
    We know that \frac{2\pi}{2}=\pi so add that to the argument of the first root to get the second square root.

    To get the cube roots divide by three instead of two. There are three roots.
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