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Math Help - complex trig function

  1. #1
    Senior Member Danneedshelp's Avatar
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    complex trig function

    Q: Solve cos(z)=\frac{3}{4}+\frac{i}{4} for z.

    My attempt:

    Let z=x+iy. Then

    cos(x+iy)=cos(x)cos(iy)-sin(x)sin(iy)

    be definition

    cos(iy)=\frac{e^{i(iy)}+e^{-i(iy)}}{2}=\frac{e^{-y}+e^{y}}{2}=-cosh(y)

    and

    sin(iy)=\frac{e^{i(iy)}-e^{-i(iy)}}{2i}=\frac{e^{-y}-e^{y}}{2i}=-\frac{sinh(y)}{i}=isinh(y)

    So, I have to solve

    -cos(x)cosh(y)=\frac{3}{4}

    and

    sin(x)sinh(y)=\frac{i}{4}

    I am not familiar with hyperbolic sine and cosine, so I am not sure what identities I can use. Is there are way to solve without using trig inverses and identities? I would think so, because our text didn't cover any examples like this.
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by Danneedshelp View Post
    Q: Solve cos(z)=\frac{3}{4}+\frac{i}{4} for z.

    My attempt:

    Let z=x+iy. Then

    cos(x+iy)=cos(x)cos(iy)-sin(x)sin(iy)

    be definition

    cos(iy)=\frac{e^{i(iy)}+e^{-i(iy)}}{2}=\frac{e^{-y}+e^{y}}{2}=-cosh(y)

    and

    sin(iy)=\frac{e^{i(iy)}-e^{-i(iy)}}{2i}=\frac{e^{-y}-e^{y}}{2i}=-\frac{sinh(y)}{i}=isinh(y)

    So, I have to solve

    -cos(x)cosh(y)=\frac{3}{4}

    and

    sin(x)sinh(y)=\frac{i}{4} Mr F says: Should be 1/4 not i/4.

    I am not familiar with hyperbolic sine and cosine, so I am not sure what identities I can use. Is there are way to solve without using trig inverses and identities? I would think so, because our text didn't cover any examples like this.
    I don't see a simple way of answering this question but a more tractable approach might be to just use the complex inverse cosine function: Inverse trigonometric functions - Wikipedia, the free encyclopedia (logarithmic forms).
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