complex no. question no. 6

• May 20th 2008, 08:27 PM
afeasfaerw23231233
complex no. question no. 6
p302 q19
given
$\displaystyle \frac{1+sinx +icosx}{1+sinx -icosx} = sin x + i cosx$
evaluate:
$\displaystyle (1+sin\frac\pi 5 +icos\frac\pi 5)^5+ i (1+sin\frac\pi 5 -icos\frac\pi 5)^5$

my working
$\displaystyle =(-i)((1+sin\frac\pi 5 +icos\frac\pi 5)^5+(1+sin\frac\pi 5 +icos\frac\pi 5)^5)$
$\displaystyle =(-i)(2(1+sin\frac\pi 5)^5-20(1+sin\frac\pi 5)^3 cos^2 \frac\pi 5 +5(1+sin \frac\pi 5) cos^4 \frac\pi 5 )$
and stuck.
thanks!
• May 21st 2008, 12:29 AM
Moo
Hello :)

Quote:

Originally Posted by afeasfaerw23231233
p302 q19
given
$\displaystyle \frac{1+sinx +icosx}{1+sinx -icosx} = sin x + i cosx$
evaluate:
$\displaystyle (1+sin\frac\pi 5 +icos\frac\pi 5)^5+ i (1+sin\frac\pi 5 -icos\frac\pi 5)^5$

my working
$\displaystyle =(-i)((1+sin\frac\pi 5 +icos\frac\pi 5)^5+(1+sin\frac\pi 5 +icos\frac\pi 5)^5)$
$\displaystyle =(-i)(2(1+sin\frac\pi 5)^5-20(1+sin\frac\pi 5)^3 cos^2 \frac\pi 5 +5(1+sin \frac\pi 5) cos^4 \frac\pi 5 )$
and stuck.
thanks!

Let :

$\displaystyle N=(1+sin\frac\pi 5 +icos\frac\pi 5)^5+ i (1+sin\frac\pi 5 -icos\frac\pi 5)^5$

Because you're given the formula above, divide by $\displaystyle (1+\sin \frac \pi 5-i \cos \frac \pi 5)^5$ (because it's different to 0, we can do it)

Therefore :

\displaystyle \begin{aligned} \frac{N}{(1+\sin \frac \pi 5-i \cos \frac \pi 5)^5} &=\frac{(1+sin\frac\pi 5 +icos\frac\pi 5)^5}{(1+\sin \frac \pi 5-i \cos \frac \pi 5)^5}+i \frac{(1+sin\frac\pi 5 -icos\frac\pi 5)^5}{(1+sin\frac\pi 5 -icos\frac\pi 5)^5} \\ &=\left(\frac{1+sin\frac\pi 5 +icos\frac\pi 5}{1+\sin \frac \pi 5-i \cos \frac \pi 5}\right)^5+i \\ &=(\sin \frac \pi 5+i \cos \frac \pi 5)^5+i \end{aligned}

Now, it would be great if it could be simplified.
You can notice that it looks like a lot de Moivre's formula :

$\displaystyle (\cos x+i \sin x)^n=\cos nx+i \sin nx$

So we're gonna make a transform... According to the unit circle (or to your lesson) :

$\displaystyle \cos (\frac \pi 2-x)=\sin x$

$\displaystyle \sin (\frac \pi 2-x)=\cos x$

Great !

Therefore $\displaystyle \sin \frac \pi 5+i \cos \frac \pi 5=\cos (\frac \pi 2-\frac \pi 5)+i \sin (\frac \pi 2-\frac \pi 5)=\cos \frac{3 \pi}{10}+i \sin \frac{3 \pi}{10}$

Now we can apply de Moivre's formula :

\displaystyle \begin{aligned} (\sin \frac \pi 5+i \cos \frac \pi 5)^5 &=\left(\cos \frac{3 \pi}{10}+i \sin \frac{3 \pi}{10}\right)^5 \\ \\ &=\cos \frac{3 \pi}{2}+i \sin \frac{3 \pi}{2} \\ &=0+i (-1) \\ &=-i \end{aligned}

-------------------

$\displaystyle \implies \boxed{\frac{N}{(1+\sin \frac \pi 5-i \cos \frac \pi 5)^5}=-i+i}$

Conclude :D
• May 21st 2008, 01:20 AM
afeasfaerw23231233
thanks again!