Complex help please (1 Viewer)

hscc · Apr 16, 2016

Hi everyone. How would we solve this?

We're given z = 2(cosø+isinø)
therefore, conjugate(1-z) = 1-2cosø+2isinø

Find the real part of 1/(1-z)

parad0xica · Apr 17, 2016

Let's realise the denominator

$\frac{1}{1-z} = \frac{1}{1-z} \frac{\overline{1-z}}{\overline{1-z}} = \frac{1-2\cos \phi}{|1-z|^2} + i \frac{2 \sin \phi}{|1-z|^2}$

Now we take the real part!

$Re \left( \frac{1}{1-z}\right) \\=\frac{1-2\cos \phi }{|1-z|^2} \\= \frac{1- 2\cos \phi}{|1-2\cos \phi - 2i \sin \phi|^2} \\= \frac{1-2 \cos \phi}{(1-2\cos \phi)^2 + (2 \sin \phi)^2} \\= \frac{1-2\cos \phi}{5 - 4 \cos \phi}$

Ambility · Apr 17, 2016

Here's an alternate way of setting it out, though it is much the same to parad0xica's, realise the denominator, simplify and separate into real and imaginary components.

$z=2(\cos\theta+i\sin\theta)\\\frac{1}{1-z}=\frac{1}{1-2\cos\theta-2i\sin\theta}\\\\$Multiplying by a fraction of unity to realise the denominator:$\\\\\frac{1}{1-2\cos\theta-2i\sin\theta}\times\frac{1-2\cos\theta+2i\sin\theta}{1-2\cos\theta+2i\sin\theta}\\=\frac{1-2\cos\theta+2i\sin\theta}{1-2\cos\theta+2i\sin\theta-2\cos\theta+4\cos^2\theta-4i\sin\theta\cos\theta-2i\sin\theta+4i\sin\theta\cos\theta+4\sin^2\theta}\\=\frac{1-2\cos\theta+2i\sin\theta}{5-4\cos\theta}\\\\$Now we can separate into real and imaginary parts:$\\\\Re(\frac{1}{1-z})=\frac{1-2\cos\theta}{5-4\cos\theta}$

braintic · Apr 17, 2016

Or more generally:

$\text{Since }z \overline z=\left|z \right|^2$

$\text{Then }\frac1{\overline z}=\frac z {\left|z\right|^2}$

$\text{So }\text{Re}\left(\frac1{\overline z}\right)=\frac{\text{Re} (z)}{\left|z\right|^2}$

$\text{From which the result easily follows}$

Complex help please (1 Viewer)

hscc

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parad0xica

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Ambility

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braintic

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