Complex help!! (1 Viewer)

hscsubjectstodo · Nov 29, 2009

where might one be asked to know 3u trig double angles in this topic... i know for one if asked to prove de moivre's theorum through induction... anything else??

tommykins · Nov 30, 2009

look through the past papers, normally question 6-7-8 will implement thse double angle rules and if so - it is within proofs.

cutemouse · Nov 30, 2009

At the top of my head...

Example:

If z=cosθ+isinθ

Prove that:

$\frac{2}{1+z}=1-itan\frac{\theta}{2}$

May as well post a solution:

$Now z=\frac{2}{1+cos\theta+isin\theta} \frac{(1+cos \theta )-isin\theta}{(1+cos \theta )-isin \theta } \\ = \frac{2+2cos\theta-2isin\theta}{(1+cos\theta)^2+sin^2\theta} \\ = \frac{2(1+cos\theta-isin\theta)}{1+2cos\theta+cos^2\theta+sin^2\theta} \\ = \frac{2(2(1+cos\theta-isin\theta}{2+2cos\theta} \\ = \frac{1+cos\theta-isin\theta}{1+cos\theta} \\ Let t=tan\frac{\theta}{2}, \ \theta \neq \pi \\ So \ z=\frac{1+ \frac{1-t^2}{1+t^2} - \frac{2it}{1+t^2}}{1+\frac{1-t^2}{1+t^2} \\ = \frac{1+t^2+1-t^2 + 2it}{1+t^2+1-t^2} \\ = \frac{2+2it}{2} \\ = 1+\frac{itan\frac{\theta}{2}}$

Bahh, I hate LateX sometimes... Will fix the code later -_-;

bachviete · Nov 30, 2009

jm01 said:
At the top of my head...

Example:

If z=cosθ+isinθ

Prove that:

$\frac{2}{1+z}=1-itan\frac{\theta}{2}$

May as well post a solution:

$$Now$\,\, z=\frac{2}{1+cos\theta+isin\theta} \frac{(1+cos \theta )-isin\theta}{(1+cos \theta )-isin \theta }$

$= \frac{2+2cos\theta-2isin\theta}{(1+cos\theta)^2+sin^2\theta}$

$= \frac{2(1+cos\theta-isin\theta)}{1+2cos\theta+cos^2\theta+sin^2\theta}$

$= \frac{2(1+cos\theta-isin\theta)}{2+2cos\theta}$

$= \frac{1+cos\theta-isin\theta}{1+cos\theta}$

$$Let$\,\, t=tan\frac{\theta}{2}, \,\, theta \neq \pi \,\,$

$$So$ \,\, \ z=\frac{1+ \frac{1-t^2}{1+t^2} - \frac{2it}{1+t^2}}{1+\frac{1-t^2}{1+t^2}}$

$= \frac{1+t^2+1-t^2 - 2it}{1+t^2+1-t^2}$

$= \frac{2-2it}{2}$

$= 1 - it$

$= 1 - i \tan\frac{\theta}{2}$
Bahh, I hate LateX sometimes... Will fix the code later -_-;

fixed

or more simply from $= \frac{1+cos\theta-isin\theta}{1+cos\theta}$

$= \frac{1+cos\theta}{1+cos\theta} - \frac{isin\theta}{1+cos\theta}$

$= 1 - \frac{isin\theta}{1+cos\theta}$

$$Let$\,\, t=tan\frac{\theta}{2}, \,\, theta \neq \pi \,\,$

$$Then$\,\, z = 1 - \frac{i\cdot \frac{2t}{1+t^2}}{1+ \frac{1-t^2}{1+t^2}}$

$= 1 - \frac{2it}{1+t^2+1-t^2}$

$=1 - itan\frac{\theta}{2}$

untouchablecuz · Nov 30, 2009

h3ll h0und · Nov 30, 2009

untouchablecuz said:
$\text{Another way to do it:}\\LHS\\=\frac{2}{1+z}\\=\frac{2}{1+cos\theta+isin\theta}\\=\frac{2}{1+(2cos^2\frac{\theta}{2}-1)+i(2sin\frac{\theta}{2}cos\frac{\theta}{2})}\\=\frac{2}{2cos^2\frac{\theta}{2}+2isin\frac{\theta}{2}cos\frac{\theta}{2}}\\=\frac{1}{cos\frac{\theta}{2}}(\frac{1}{cos\frac{\theta}{2}+isin\frac{\theta}{2}})\\=\frac{1}{cos\frac{\theta}{2}}(cos\frac{\theta}{2}-isin\frac{\theta}{2})\\=1-itan\frac{\theta}{2}\\=RHS$

how did u get from the 5th to 6th step?
is 1/cos theta/2 = cos -theta/2? and 1/sin theta/2 = sin -theta/2 ?

khorne · Nov 30, 2009

h3ll h0und said:
how did u get from the 5th to 6th step?
is 1/cos theta/2 = cos -theta/2? and 1/sin theta/2 = sin -theta/2 ?

multiply top/bottom by cos - isin@/2, and on bottom, you have cos^2 @ + sin^2 @ = 1, thus it's just cos@ - isin@

untouchablecuz · Nov 30, 2009

h3ll h0und said:
how did u get from the 5th to 6th step?
is 1/cos theta/2 = cos -theta/2? and 1/sin theta/2 = sin -theta/2 ?

$z\bar{z}=|z|^2\Rightarrow \frac{|z|^2}{z}=\bar{z}\\\text{In this case:}\\z=cos\frac{\theta}{2}+isin\frac{\theta}{2}, \\\bar{z}=cos(-\frac{\theta}{2})+isin(-\frac{\theta}{2})=cos\frac{\theta}{2}-isin\frac{\theta}{2}\\|z|=1\\\therefore \frac{1}{cos\frac{\theta}{2}+isin\frac{\theta}{2}}=cos\frac{\theta}{2}-isin\frac{\theta}{2}\\$

Complex help!! (1 Viewer)

hscsubjectstodo

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