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Math Help - z = re^i(theta)

  1. #1
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    z = re^i(theta)

    I have noooooo idea how to do these types of problems. I missed the lecture when we were on this topic and now I'm lost. Can anyone explain to me how to do these type, step by step?

    1. Express the complex number in polar form:

    a) (e^i*pi/3)^2

    b) (2+3i) + (-5-7i)

    c) 4root(10e^i*pi/2)

    d) (2i)^3 - (2i)^2 + 2i - 1

    Please help...
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  2. #2
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    Quote Originally Posted by ohhhh View Post
    1. Express the complex number in polar form:

    a) (e^i*pi/3)^2
    (e^{i \pi/3})^2=e^{i 2 \pi/3}

    which may be what you want of in (r, \theta) form this is (1, 2 \pi/3)

    RonL
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  3. #3
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    Quote Originally Posted by ohhhh View Post
    1. Express the complex number in polar form:

    b) (2+3i) + (-5-7i)
     <br />
(2+3i) + (-5-7i) = -3 -4i = \sqrt{3^2+4^2}(\cos(\theta)+i\sin(\theta))=5e^{i\t  heta}<br />

    where \theta=-\arctan(4/3)

    RonL
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  4. #4
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    what about a problem like (-1-i)^4/3? how do i "find the quantity" and then change the form to cartesian?
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  5. #5
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    Quote Originally Posted by ohhhh View Post
    what about a problem like (-1-i)^4/3? how do i "find the quantity" and then change the form to cartesian?
     <br />
-1-i =\sqrt{2} e^{\frac{3 \pi i}{2}+2 \pi i n}, \ n = 0, \pm1, ..<br />

    so:

     <br />
(-1-i)^{4/3} =4^{1/3} e^{\frac{4 \pi i}{2}+8 \pi i n/3}=4^{1/3} e^{8 \pi i n/3}, \ n = 0, \pm1, ..<br />

    of which only three values are distinct:

     <br />
(-1-i)^{4/3} =4^{1/3} e^{8 \pi i n/3}, \ n = 0, 1, 2 <br />

    RonL
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