Express the eq in a+ib form

**TomJerry** · January 4th 2010, 04:47 PM

Express $z = \frac{(cos \theta + i sin \theta)^3(sin \theta + i cos \theta)}{(cos2 \theta - i sin2 \theta)}$ in the form $a+ib$

Solution :

$z = \frac{(cos \theta + i sin \theta)^3(sin \theta + i cos \theta)}{(cos2 \theta - i sin2 \theta)}$

$= \frac{(cos \theta + i sin \theta)^3(sin \theta + i cos \theta)}{(cos \theta + i sin \theta)^{-2}}$

$= (cos \theta + i sin \theta)^5(sin \theta + i cos \theta)$

Stuck !!!!!

**mr fantastic** · January 4th 2010, 05:20 PM

Originally Posted by TomJerry

Express $z = \frac{(cos \theta + i sin \theta)^3(sin \theta + i cos \theta)}{(cos2 \theta - i sin2 \theta)}$ in the form $a+ib$

Solution :

$z = \frac{(cos \theta + i sin \theta)^3(sin \theta + i cos \theta)}{(cos2 \theta - i sin2 \theta)}$

$= \frac{(cos \theta + i sin \theta)^3(sin \theta + i cos \theta)}{(cos \theta + i sin \theta)^{-2}}$

$= (cos \theta + i sin \theta)^5(sin \theta + i cos \theta)$

Stuck !!!!!

Assuming your last line is correct (and I haven't bothered to check) then you can use deMoivre's Theorem to write $( \cos \theta + i \sin \theta )^5 = \cos (5 \theta) + i \sin (5 \theta)$ . Now expand $(\cos (5 \theta) + i \sin (5 \theta)) (\cos ( \theta) + i \sin (\theta))$ and group the real and imaginary parts.

**TomJerry** · January 4th 2010, 05:48 PM

Originally Posted by mr fantastic

Assuming your last line is correct (and I haven't bothered to check) then you can use deMoivre's Theorem to write $( \cos \theta + i \sin \theta )^5 = \cos (5 \theta) + i \sin (5 \theta)$ . Now expand $(\cos (5 \theta) + i \sin (5 \theta)) (\cos ( \theta) + i \sin (\theta))$ and group the real and imaginary parts.

U are missing the question if u look closely the last line is

$(cos 5\theta + i sin 5\theta) (sin \theta + i cos \theta)$

and not as u have mentioned it

$(cos 5 \theta + i sin 5\theta) (cos \theta + i sin \theta)$

**mr fantastic** · January 4th 2010, 06:01 PM

Originally Posted by TomJerry

U are missing the question if u look closely the last line is

$(cos 5\theta + i sin 5\theta) (sin \theta + i cos \theta)$

and not as u have mentioned it

$(cos 5 \theta + i sin 5\theta) (cos \theta + i sin \theta)$

You should realise that my small error has absolutely no affect on the advice I have given you.

**Pulock2009** · January 4th 2010, 08:59 PM

sin (theta)=cos((pi)/2-(theta))
cos(theta)=sin((pi)/2-(theta))

**TomJerry** · January 4th 2010, 09:06 PM

Originally Posted by mr fantastic

You should realise that my small error has absolutely no affect on the advice I have given you.

how will u group the real parts when the one is a cos and the other is a sin ..........

**mr fantastic** · January 4th 2010, 09:44 PM

Originally Posted by TomJerry

how will u group the real parts when the one is a cos and the other is a sin ..........

Can you not expand $(\cos (5\theta) + i \sin (5\theta)) (\sin \theta + i \cos \theta)$ and then group together the real and imaginary bits??

**TomJerry** · January 5th 2010, 01:40 AM

Originally Posted by mr fantastic

Can you not expand $(\cos (5\theta) + i \sin (5\theta)) (\sin \theta + i \cos \theta)$ and then group together the real and imaginary bits??

could u please show me a example or even a link would be more help.......i just need to know how to carry out this .......different signs

**Prove It** · January 5th 2010, 02:27 AM

Originally Posted by TomJerry

could u please show me a example or even a link would be more help.......i just need to know how to carry out this .......different signs

$(\cos{5\theta} + i\sin{5\theta})(\sin{\theta} + i\cos{\theta}) = \cos{5\theta}\sin{\theta} + i\cos{5\theta}\cos{\theta} + i\sin{5\theta}\sin{\theta} + i^2\sin{5\theta}\cos{\theta}$

$= \cos{5\theta}\sin{\theta} - \sin{5\theta}\sin{\theta} + i(\cos{5\theta}\cos{\theta} + \sin{5\theta}\sin{\theta})$ .

What do you think $a$ and $b$ are?

**Plato** · January 5th 2010, 07:10 AM

Here are some short cuts.
$\frac{{\left[ {\cos (\varphi ) + i\sin (\varphi )} \right]^3 \left( {\sin (\varphi ) + i\cos (\varphi )} \right)}}{{\cos (2\varphi ) - i\sin (2\varphi )}}$ $= \left[ {\cos (3\varphi ) + i\sin (3\varphi )} \right]\left( {\sin (\varphi ) + i\cos (\varphi )} \right)\left( {\cos (2\varphi ) + i\sin (2\varphi )} \right)$
$\left( {\sin (\varphi ) + i\cos (\varphi )} \right) = \left( {\cos (\varphi ) - i\sin (\varphi )} \right)i = \left( {\cos ( - \varphi ) + i\sin ( - \varphi )} \right)i$

**TomJerry** · January 6th 2010, 01:07 AM

thanks

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