Complex Numbers Help (1 Viewer)

Rono123 · Nov 5, 2011

Prove the following:

(1 - cosx + 2isinx)^-1 = (1 - 2icot(x/2))/(5 + 3cosx)

Thanks!!

seanieg89 · Nov 5, 2011

SpiralFlex · Nov 5, 2011

$LHS=\frac{1}{1-\cos x +2i\sin x}*\frac{1-\cos x-2i\sin x}{1-\cos x-2i\sin x}$

$=\frac{1-\cos x-2i\sin x}{(1-\cos x)^2 + 4\sin^2 x}$

$=\frac{1-\cos x-2i\sin x}{1-2\cos x + \cos^2 x + 4(1-\cos^2 x)}$

$=\frac{1-\cos x-2i\sin x}{-3\cos^2 x-2\cos x + 5}$

$=\frac{1-\cos x-2i\sin x}{-(3\cos x + 5)(\cos x - 1)}$

Remember our double angles,

$1-\cos x = 2\sin^2 \frac{x}{2}$

Applying this,

$=\frac{2\sin^2 \frac{x}{2}-4i\sin \frac{x}{2} \cos \frac{x}{2}}{-(3\cos x + 5)*-2\sin^2 \frac{x}{2}}$

$=\frac{2\sin \frac{x}{2}(\sin \frac{x}{2}-2i\cos \frac{x}{2})}{-(3\cos x + 5)*-2\sin^2 \frac{x}{2}}$

$=\frac{(\sin \frac{x}{2}-2i\cos \frac{x}{2})}{(3\cos x + 5)*\sin \frac{x}{2}}$

$=\frac{1-2i \cot \frac{x}{2}}{5+3\cos x}$

$=RHS$

Rono123 · Nov 5, 2011

Thank you soo much!! I've been puzzling over this for hours....so thanks a lot!

math man · Nov 6, 2011

Another method if you are interested is to first convert into half angles:

$LHS= (1-(1-2sin^{2}\frac{x}{2})+4isin\frac{x}{2}cos\frac{x}{2})^{-1}$

this simplifies to:

$LHS= (2sin\frac{x}{2}(sin\frac{x}{2}+2icos\frac{x}{2}))^{-1}$

using conjugate properties this becomes:

$LHS= \frac{1}{2sin\frac{x}{2}}\frac{sin\frac{x}{2}-2icos\frac{x}{2}}{sin^{2}\frac{x}{2}+4cos^{2}\frac{x}{2}}$

dividing everything by sinx/2 and subbing in the double angles for cosx/2 gives:

$LHS= \frac{1}{2}\frac{1-2icot\frac{x}{2}}{1+3(\frac{1}{2}(cosx+1))}$

which simplifies to:

$LHS= \frac{1-2icot\frac{x}{2}}{5+3cosx}=RHS$

note in the previous step i broke sin^2(x/2) + 4cos^2(x/2) into sin^2(x/2) + cos^2(x/2) + 3cos^2(x/2) so i could take one out

Complex Numbers Help (1 Viewer)

Rono123

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seanieg89

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SpiralFlex

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Rono123

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math man

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