# complex trig function

• Sep 22nd 2010, 10:04 PM
Danneedshelp
complex trig function
Q: Solve $\displaystyle cos(z)=\frac{3}{4}+\frac{i}{4}$ for $\displaystyle z$.

My attempt:

Let $\displaystyle z=x+iy$. Then

$\displaystyle cos(x+iy)=cos(x)cos(iy)-sin(x)sin(iy)$

be definition

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

and

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

So, I have to solve

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

and

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

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.
• Sep 23rd 2010, 12:33 AM
mr fantastic
Quote:

Originally Posted by Danneedshelp
Q: Solve $\displaystyle cos(z)=\frac{3}{4}+\frac{i}{4}$ for $\displaystyle z$.

My attempt:

Let $\displaystyle z=x+iy$. Then

$\displaystyle cos(x+iy)=cos(x)cos(iy)-sin(x)sin(iy)$

be definition

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

and

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

So, I have to solve

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

and

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