# [SOLVED] proof for complx/trig eqn?

• Jan 7th 2009, 11:23 AM
crb
[SOLVED] proof for complx/trig eqn?
plz have a look at the aatached picture
thank you
• Jan 7th 2009, 11:30 AM
Mush
$\displaystyle sinh(z) = \frac{e^{z} - e^{-z}}{2}$

$\displaystyle = \frac{e^{x+iy} - e^{-x-iy}}{2}$

$\displaystyle = \frac{e^{x}.e^{iy} - e^{-x}.e^{-iy}}{2}$

$\displaystyle = \frac{e^{x}.e^{iy} - e^{-x}.\frac{1}{e^{iy}}}{2}$

$\displaystyle = \frac{e^{x}.(cos(y)+isin(y)) - \frac{1}{e^{x}.(cos(y)+isin(y))}}{2}$
• Jan 7th 2009, 11:41 AM
crb
and how about the conjugate of sinh(z)....??
how do i get sinh(z) = some u + iv form????
• Jan 7th 2009, 11:44 AM
Mush
Quote:

Originally Posted by crb
and how about the conjugate of sinh(z)....??
how do i get sinh(z) = some u + iv form????

$\displaystyle = \frac{e^{x}.(cos(y)+isin(y)) - \frac{1}{e^{x}.(cos(y)+isin(y))}}{2}$

Here you have a complex number which is composed of subtraction of two complex numbers.

Remember that if $\displaystyle w$ and $\displaystyle z$ are complex numbers then:

$\displaystyle \overline{(v - w)} = \overline{v} - \overline{w}$

In this case $\displaystyle v = e^{x}.cos(y)+ie^{x}sin(y)$ $\displaystyle w = \frac{1}{e^{x}.cos(y)+ie^xsin(y)}$

And you're trying to find $\displaystyle \frac{1}{2} \overline{(v - w)}$

Get $\displaystyle w$ in a nicer form by multiplying top and bottom by the conjugate of the denominator first.

$\displaystyle w = \frac{1}{e^{x}.cos(y)+ie^xsin(y)} = \frac{e^{x}.cos(y)-ie^xsin(y)}{e^{2x}.cos^2(y)+ie^{2x}sin^2(y)} = \frac{cos(y)-isin(y)}{e^{x}}$

Hence

$\displaystyle \frac{1}{2}( v - w) = \frac{1}{2}(e^{x}.cos(y)+ie^{x}sin(y) - (\frac{cos(y)-isin(y)}{e^{x}})$

Should be relatively easy to find the conjugate now.
• Jan 7th 2009, 12:00 PM
crb
Thank u very much...just managed to prove it with ur help..thank you