1. ## 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)

$
\frac{e^{ix-y} - e^{-ix+y}}{2i}
$

RHS
sin x cosh y + i cos x sinh y

$
\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}
$

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

$
\frac{8e^{ix-y} + 8e^{-ix+y}}{16i}
$

Which is equal to
$
\frac{e^{ix-y} - e^{-ix+y}}{2i}
$

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?

2. Originally Posted by Richmond
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)

$
\frac{e^{ix-y} - e^{-ix+y}}{2i}
$

RHS
sin x cosh y + i cos x sinh y

$
\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}
$

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

$
\frac{8e^{ix-y} + 8e^{-ix+y}}{16i}
$

Which is equal to
$
\frac{e^{ix-y} - e^{-ix+y}}{2i}
$

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 $x$.

3. Thanks. sorry to bother with one last part

part c)

Show that $x = 2n\pi+\frac{\pi}{2}$ for some integer n and $y = \pm \log(2+\sqrt{3})$

4. Originally Posted by Richmond
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.

5. Thanks alot. I got it.