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Math Help - function values of complex functions

  1. #1
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    function values of complex functions

    This problem says:

    Showing the steps, not just giving the final answer, find the values of:

    a. e^(4 + 2i)

    b. Ln(5 - 2i)

    c. sin (4(pi) - (pi)i/2)

    d. (1 + i)^(2i)

    This stuff will be on my final on Tuesday, and I don't really get how to do it at all. If someone would help me with these, that would be great. Please show me how you worked the problem so that I will be able to understand this stuff for the test. Thanks!
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  2. #2
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    Quote Originally Posted by Hollysti View Post
    a. e^(4 + 2i)
    This one's easy.

    e^{4 + 2i} = e^4 * e^{2i}

    = e^4 * (cos(2) + i*sin(2))

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Hollysti View Post
    c. sin (4(pi) - (pi)i/2)
    This one is easy if you recall that even for the complex extension of the sine function we still have:
    sin(A + B) = sin(a)cos(B) + sin(B)cos(A)

    So:
    sin(4(pi) - i*(pi)/2) = sin(4(pi))cos(i*(pi)/2) - sin(i*(pi)/2)cos(4(pi))

    = -sin(i*(pi)/2)

    Now
    sin(x) = (1/(2i))*(e^{ix} - e^{-ix})

    So
    sin(i*(pi)/2) = (1/(2i))*(e^{i*i*(pi)/2} - e^{-i*i*(pi)/2})

    = (1/(2i))*(e^{-(pi)/2} - e^{(pi)/2})

    = (1/i)*-sinh((pi)/2) = i*sinh((pi)/2)

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Hollysti View Post
    b. Ln(5 - 2i)
    The function is "ln" by the way, not "Ln."

    ln(z) = ln|z| + i*arg(z)

    So
    ln(5 - 2i) = ln|5 - 2i| + i*arg(5 - 2i)

    Let's convert 5 - 2i to complex polar form:
    5 - 2i = re^{it} (t being shorthand for "theta.")

    r = sqrt{5^2 + (-2)^2} = sqrt{29}
    t = atn{-2/5} = -atn{2/5}

    So
    5 - 2i = sqrt{29}*e^{-i*atn(2/5)}

    Thus
    ln|5 - 2i| = ln(sqrt{29}) = (1/2)*ln(29)
    arg(5 - 2i) = -atn(2/5)

    So
    ln(5 - 2i) = (1/2)*ln(29) - i*atn(2/5)

    -Dan
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Hollysti View Post
    d. (1 + i)^(2i)
    Let x = (1 + i)^{2i}

    Then
    ln(x) = (2i)*ln(1 + i)

    ln(1 + i) = ln|1 + i| + i*arg(1 + i)

    = (1/2)*ln(2) + i*atn(1) = (1/2)*ln(2) + i*(pi)/4

    So
    ln(x) = (2i) * ((1/2)ln(2) + i*(pi)/4)

    = i*ln(2) - (pi)/2

    = -(pi)/2 + i*ln(2)

    Thus
    x = e^{-(pi)/2 + i*ln(2)} = [e^{-(pi)/2}] * [e^{i*ln(2)}]

    = [e^{-(pi)/2}] * [cos(ln(2)) + i*sin(ln(2))]

    = [e^{-(pi)/2}]*[cos(ln(2)) + i*sin(ln(2))]

    -Dan
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