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Math Help - Complex equation

  1. #1
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    Complex equation

    Solve: e^{z^2}=i

    I tried writing it like this:  e^{z^2} = e^{e^{2\log(z)}}=e^{\pi i /2+2\pi i n

    That leaves us with: e^{2\log(z)} = \pi i /2+2\pi i n, but that didn't help much.
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  2. #2
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    use the identity  i = e^{\frac{i \pi}{2} + 2 \pi i k}, k = 0, \pm 1, \pm 2 , \ldots
    Last edited by Defunkt; June 20th 2010 at 02:57 AM. Reason: mistake
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  3. #3
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    I've already used it, and it leads me to the last equation.
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  4. #4
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    Quote Originally Posted by Mondreus View Post
    Solve: e^{z^2}=i

    I tried writing it like this:  e^{z^2} = e^{e^{2\log(z)}}=e^{\pi i /2+2\pi i n

    That leaves us with: e^{2\log(z)} = \pi i /2+2\pi i n, but that didn't help much.
    e^{z^2} = i

    e^{(x + iy)^2} = i

    e^{x^2 + 2ixy +i^2y^2} = i

    e^{x^2 - y^2 + 2ixy} = i

    e^{x^2 - y^2}\,\textrm{cis}\,(2xy) = i

    e^{x^2 - y^2}\cos{(2xy)} + i\,e^{x^2 - y^2}\sin{(2xy)} = 0 + i.


    Equating real and imaginary parts gives:

    e^{x^2 - y^2}\cos{(2xy)} = 0 and e^{x^2 - y^2}\sin{(2xy)} = 1.


    Equation 1:

    e^{x^2 - y^2}\cos{(2xy)} = 0

    Since the exponential function is always positive, this means

    \cos{(2xy)} = 0.


    From the Pythagorean Identity:

    \cos^2{(2xy)} + \sin^2{(2xy)} = 1

    \sin^2{(2xy)} = 1 since \cos{(2xy)} = 0.


    Can you go from here?
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  5. #5
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    The answer is z=\pm(1+i)\sqrt{\pi/4+n\pi} and z=\pm(1-i)\sqrt{-\pi/4+n\pi}, but I can't there from what you posted.
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  6. #6
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    Quote Originally Posted by Prove It View Post
    e^{z^2} = i

    e^{(x + iy)^2} = i

    e^{x^2 + 2ixy +i^2y^2} = i

    e^{x^2 - y^2 + 2ixy} = i

    e^{x^2 - y^2}\,\textrm{cis}\,(2xy) = i

    e^{x^2 - y^2}\cos{(2xy)} + i\,e^{x^2 - y^2}\sin{(2xy)} = 0 + i.


    Equating real and imaginary parts gives:

    e^{x^2 - y^2}\cos{(2xy)} = 0 and e^{x^2 - y^2}\sin{(2xy)} = 1.


    Equation 1:

    e^{x^2 - y^2}\cos{(2xy)} = 0

    Since the exponential function is always positive, this means

    \cos{(2xy)} = 0.


    From the Pythagorean Identity:

    \cos^2{(2xy)} + \sin^2{(2xy)} = 1

    \sin^2{(2xy)} = 1 since \cos{(2xy)} = 0.


    Can you go from here?
    Well, from where I left off...

    \sin^2{(2xy)} = 1 since \cos{(2xy)} = 0

    \sin{(2xy)} = 1 or \sin{(2xy)} = -1.


    Substituting into equation 2:

    e^{x^2 - y^2}\sin{(2xy)} = 1

    Since e^{x^2 - y^2} is always positive, that means the only acceptable solution is \sin{(2xy)} = 1.

    So e^{x^2 - y^2} = 1

    x^2 - y^2 = \ln{1}

    x^2 - y^2 = 0

    y^2 = x^2

    y = \pm x.


    Subtituting back into \cos{(2xy)} = 0

    \cos{[2x(\pm x)]} = 0

    \cos{(\pm 2x^2)} = 0

    \pm 2x^2 = \frac{\pi}{2} + \pi n where n \in \mathbf{Z}

    x^2 = \pm\left(\frac{\pi}{4} + \frac{\pi n}{2}\right)

    x = \pm \sqrt{ \pm(\frac{\pi}{4} + \frac{\pi n}{2} \right ) }.


    Since y = \pm that means y = \pm \sqrt{\pm \left(\frac{\pi}{4} + \frac{\pi n}{2}\right)}.


    Can you go from here?
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  7. #7
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    That seems to be more solutions than what the textbook says though. This is how I "solved" it, but I'm not sure everything I've done is correct, and I seem to be missing some solutions. I'm especially confused about taking the square root of a complex number, and when you can factor out complex numbers from square roots.

    z^2 = i(\pi /2 + 2 \pi  n) \Leftrightarrow z^2 = 2e^{i(\pi /2 + 2 \pi  k)}(\pi / 4 +  \pi n) \Leftrightarrow

    z = \pm \sqrt{2}\sqrt{e^{i(\pi /2 + 2 \pi  k)}(\pi / 4 +  \pi n)} = \pm \sqrt{2}\sqrt{e^{i(\pi /2 + 2 \pi  k)}}\sqrt{\pi / 4 +  \pi n} = \pm(1+i)\sqrt{\pi / 4 +  \pi n}

    I know that this isn't true in general if z and w are complex numbers: \sqrt{z}\sqrt{w} = \sqrt{zw}. It's often useful to factor out -1 = i^2 from square roots though, so how do you know when you can do it?
    Last edited by Mondreus; June 21st 2010 at 01:56 AM.
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