# Here is my second question on the complex numbers assignment. How do I prove the following below?

## If z=cos$\theta$+isin$\theta$, prove that; 1. 1+$z + {z}^{2}$=(1+cos theta)(cos theta+isin theta) 2. $z / \left(1 + z\right) = 1 + i \tan \left(\theta / 2\right)$

Aug 10, 2018

See below

#### Explanation:

1. $z = \cos \theta + i \sin \theta$

$z + 1 = \cos \theta + i \sin \theta + 1$

${\left(z + 1\right)}^{2} = {\left(\cos \theta + i \sin \theta + 1\right)}^{2}$

$1 + 2 z + {z}^{2} = {\cos}^{2} \theta + 2 \cos \theta i \sin \theta + 2 \cos \theta + {i}^{2} {\sin}^{2} \theta + 2 i \sin \theta + 1$

$1 + 2 z + {z}^{2} = {\cos}^{2} \theta + 2 \cos \theta i \sin \theta + 2 \cos \theta - {\sin}^{2} \theta + 2 i \sin \theta + {\sin}^{2} \theta + {\cos}^{2} \theta$

$1 + 2 z + {z}^{2} = 2 {\cos}^{2} \theta + 2 \cos \theta i \sin \theta + 2 \cos \theta + 2 i \sin \theta$

Factor:
$1 + 2 z + {z}^{2} = 2 \left(1 + \cos \theta\right) \left(\cos \theta + i \sin \theta\right)$

Aug 10, 2018

You asked this:

For 1), you are proving that:

• $1 + z + {z}^{2} = z \left(1 + \cos \theta\right) q \quad \equiv q \quad \frac{1}{z} + 1 + z = 1 + \cos \theta$

$z$ is the unit circle so I don't see a problem with dividing like that.

Well:

$q \quad \frac{1}{z} = \frac{\overline{z}}{z \overline{z}} = \frac{\cos \theta - i \sin \theta}{{\cos}^{2} \theta + {\sin}^{2} \theta} = \cos \theta - i \sin \theta$

So:

$q \quad \frac{1}{z} + 1 + z = \cos \theta - i \sin \theta + 1 + \cos \theta + i \sin \theta$

$q \quad = 2 \cos \theta + 1$

$\implies 1 + z + {z}^{2} = \left(1 + \boldsymbol{2} \cos \theta\right) \left(\cos \theta + i \sin \theta\right)$

That's not what you're looking for but it is the same as other answer posted here for this question, if you actually finish off the algebra.

For 2) , I think the answer is out by a factor of 2:

$\frac{z}{1 + z} = \frac{1 + z - 1}{1 + z}$

$= 1 - \frac{1}{1 + z}$

$= 1 - \frac{\overline{1 + z}}{\left(1 + z\right) \overline{\left(1 + z\right)}}$

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

$= 1 - \frac{1 + \cos \theta - i \sin \theta}{2 + 2 \cos \theta}$

$= \frac{1}{2} + i \frac{\sin \theta}{2 + 2 \cos \theta}$

Half angle formulae:

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

$= \frac{1}{2} + i \frac{2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}{4 {\cos}^{2} \left(\frac{\theta}{2}\right)}$

$= \boldsymbol{\frac{1}{2}} \left(1 + i \tan \left(\frac{\theta}{2}\right)\right)$

Again not the answer you're looking for but I don't see the mistake in the algebra.