# A few complex number qs (1 Viewer)

#### 5uckerberg

##### Well-Known Member
For d when you have $\bg_white \arg\left(z^{2}\right)$ this implies that you need to double the argument to obtain the arg of z so in turn

$\bg_white \arg\left(z^{2}\right)=2\arg\left(z\right)=\frac{2\pi}{3}$.
Thus, $\bg_white \arg\left(z\right)=\frac{\pi}{3}$

#### =)(=

##### Active Member
I got that but in the solution for some reason they also drew in (-2pi/3)

#### yanujw

##### Active Member
I got that but in the solution for some reason they also drew in (-2pi/3)
This is because squaring complex nos with arg (-2pi/3) gives them an arg of -4pi/3, which is an equilvalent arg to 2pi/3 and therefore satisfies the condition.

#### =)(=

##### Active Member
ohh okay I see but how would you get that working from pi/3

#### 5uckerberg

##### Well-Known Member
De Moivre's Theorem that is what is happening squaring the angle doubles the angle of rotation and so on.

#### =)(=

##### Active Member
I still can't see it for some reason

#### 5uckerberg

##### Well-Known Member
Part b well we need to know this fact $\bg_white \omega\bar{\omega}=|\omega|^{2}=1$ and $\bg_white \omega+\bar{\omega}=2Re\left(\omega\right)$ and $\bg_white \omega-\bar{\omega}=2Im\left(\omega\right)$
Once this is known our next step is pretty straightforward.

$\bg_white \frac{1+\omega}{1-\omega}=\frac{\left(1+\omega\right)\left(1-\bar{\omega}\right)}{\left(1-\omega\right)\left(1-\bar{\omega}\right)}$.

$\bg_white \frac{1+\omega-\bar{\omega}-\omega\bar{\omega}}{1-\omega-\bar{\omega}+\omega\bar{\omega}}$.

$\bg_white \frac{2Im\left(\omega\right)}{2-2Re\left(\omega\right)}$

$\bg_white \frac{Im\left(\omega\right)}{1-Re\left(\omega\right)}$

$\bg_white \frac{iy}{1-x}$

I would like @=)(= to confirm the answer for this one.

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#### =)(=

##### Active Member
Part b well we need to know this fact $\bg_white \omega\bar{\omega}=|\omega|^{2}=1$ and $\bg_white \omega+\bar{\omega}=2Re\left(\omega\right)$ and $\bg_white \omega-\bar{\omega}=2Im\left(\omega\right)$
Once this is known our next step is pretty straightforward.

$\bg_white \frac{1+\omega}{1-\omega}=\frac{\left(1+\omega\right)\left(1-\bar{\omega}\right)}{\left(1-\omega\right)\left(1-\bar{\omega}\right)}$.

$\bg_white \frac{1+\omega-\bar{\omega}-\omega\bar{\omega}}{1-\omega-\bar{\omega}+\omega\bar{\omega}}$.

$\bg_white \frac{2Im\left(\omega\right)}{2-2Re\left(\omega\right)}$

$\bg_white \frac{Im\left(\omega\right)}{1-Re\left(\omega\right)}$

$\bg_white \frac{y}{1-x}$

I would like @=)(= to confirm the answer for this one.