# Math Help - complex trig function

1. ## 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.

2. Originally Posted by Danneedshelp
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).