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Math Help - Complex exponential

  1. #1
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    Complex exponential

    Show that

    part a)
    sin(x+iy) = sin x cosh y+ i cos x sinh y
    using complex exponentials

    This is what i got so far

    LHS
    sin(x+iy)

    <br />
\frac{e^{ix-y} - e^{-ix+y}}{2i}<br />

    RHS
    sin x cosh y + i cos x sinh y

    <br />
\frac{e^{ix+y} + e^{ix-y} - e^{-ix+y} - e^{-ix-y}}{4i} + i\frac{e^{ix+y} - e^{ix-y} + e^{-ix+y} - e^{-ix-y}}{4}<br />

    when i combine(cross multiply) both the RHS equation i get

    <br />
\frac{8e^{ix-y} + 8e^{-ix+y}}{16i}<br />

    Which is equal to
    <br />
\frac{e^{ix-y} - e^{-ix+y}}{2i}<br />

    LHS = RHS

    part b)
    If z = x+iy with x and y real and if sin(z) = 2, deduce that cos x = 0
    how do I go upon that?
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  2. #2
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    Quote Originally Posted by Richmond View Post
    Show that

    part a)
    sin(x+iy) = sin x cosh y+ i cos x sinh y
    using complex exponentials

    This is what i got so far

    LHS
    sin(x+iy)

    <br />
\frac{e^{ix-y} - e^{-ix+y}}{2i}<br />

    RHS
    sin x cosh y + i cos x sinh y

    <br />
\frac{e^{ix+y} + e^{ix-y} - e^{-ix+y} - e^{-ix-y}}{4i} + i\frac{e^{ix+y} - e^{ix-y} + e^{-ix+y} - e^{-ix-y}}{4}<br />

    when i combine(cross multiply) both the RHS equation i get

    <br />
\frac{8e^{ix-y} + 8e^{-ix+y}}{16i}<br />

    Which is equal to
    <br />
\frac{e^{ix-y} - e^{-ix+y}}{2i}<br />

    LHS = RHS

    part b)
    If z = x+iy with x and y real and if sin(z) = 2, deduce that cos x = 0
    how do I go upon that?
    You know that \sin (z) = \sin x \cosh y + i \cos x \sinh y where z = x + iy.

    Therefore 2 = \sin x \cosh y + i \cos x \sinh y and so equating real and imaginary parts gives

    2 = \sin x \cosh y .... (1)

    0 = \cos x \sinh y .... (2)

    From (2) either \cos x = 0 or \sinh y = 0.

    But if \sinh y = 0 then \cosh y = \pm 1. From (1) it would therefore follow that \sin x = \pm 2 which has no real solution for x.
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  3. #3
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    Thanks. sorry to bother with one last part

    part c)

    Show that x = 2n\pi+\frac{\pi}{2} for some integer n and y = \pm \log(2+\sqrt{3})
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  4. #4
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    Quote Originally Posted by Richmond View Post
    Thanks. sorry to bother with one last part

    part c)

    Show that x = 2n\pi+\frac{\pi}{2} for some integer n and y = \pm \log(2+\sqrt{3})
    You should be able to solve \cos x = 0 for x.

    \cos x = 0 \Rightarrow \sin x = \pm 1. From (1), \sin x = -1 \Rightarrow \cosh y = -2 which has no real solution.

    Therefore \sin x = 1 and so from (1) \cosh y = 2.

    \cosh y = 2 \Rightarrow \frac{e^y + e^{-y}}{2} = 2 \Rightarrow e^y + e^{-y} = 4.

    The last bit of the solution for y is left for you to attempt.
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  5. #5
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    Thanks alot. I got it.
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