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)

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

RHS
sin x cosh y + i cos x sinh y

$\displaystyle \frac{e^{ix+y} + e^{ix-y} - e^{-ix+y} - e^{-ix-y}}{4i} + i\frac{e^{ix+y} - e^{ix-y} + e^{-ix+y} - e^{-ix-y}}{4}$

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

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

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

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)

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

RHS
sin x cosh y + i cos x sinh y

$\displaystyle \frac{e^{ix+y} + e^{ix-y} - e^{-ix+y} - e^{-ix-y}}{4i} + i\frac{e^{ix+y} - e^{ix-y} + e^{-ix+y} - e^{-ix-y}}{4}$

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

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

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

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 $\displaystyle \sin (z) = \sin x \cosh y + i \cos x \sinh y$ where $\displaystyle z = x + iy$.

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

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

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

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

But if $\displaystyle \sinh y = 0$ then $\displaystyle \cosh y = \pm 1$. From (1) it would therefore follow that $\displaystyle \sin x = \pm 2$ which has no real solution for $\displaystyle x$.

3. Thanks. sorry to bother with one last part

part c)

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

4. Originally Posted by Richmond
Thanks. sorry to bother with one last part

part c)

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

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

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

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

The last bit of the solution for $\displaystyle y$ is left for you to attempt.

5. Thanks alot. I got it.