Stuck on Complex Numbers Question (1 Viewer)

Ambility · Nov 28, 2015

If someone could do a worked solution for this, I'd appreciate it.

$$If $|z|=r$ and $arg\: z=\theta$, show that $\frac{z}{z^2+r^2}$ is real and give its value.$

leehuan · Nov 28, 2015

Sent from my iPhone using Tapatalk

InteGrand · Nov 28, 2015

Ambility said:
If someone could do a worked solution for this, I'd appreciate it.

$$If $|z|=r$ and $arg\: z=\theta$, show that $\frac{z}{z^2+r^2}$ is real and give its value.$

$\textbf{ALTERNATE METHOD}$

$Since $\arg (z)$ is defined, we can assume $z\neq 0$. We will use the fact that a complex number $\omega \neq 0$ is real if and only if its reciprocal $\frac{1}{\omega}$ is real (this is easy to prove using polar form).$

$Denoting $\omega:= \frac{z}{z^2 + r^2}$, where $z=r\, \mathrm{cis}\left( \theta \right)$, we have$

$\begin{align*}\frac{1}{\omega} &= \frac{z^2 + r^2}{z} \\ &= z + \frac{r^2}{z} \\ &= z + \frac{r^2}{r \, \mathrm{cis}\left( \theta \right)} \\ &= z + \frac{r}{\mathrm{cis}\left( \theta \right)}\\ &= z + r \,\mathrm{cis} \left(- \theta \right) \quad \Big{(}\text{since } \frac{1}{\mathrm{cis} (\theta)} = \left( \mathrm{cis} (\theta )\right) ^{-1} = \mathrm{cis} \left(- \theta \right) \Big{)} \\ &= z + \bar{z} \\ &= 2\, \mathrm{Re}\left(z\right), \end{align*}$

$which is real and non-zero provided that $z$ is not purely imaginary, so the quantity $\omega$ is also real, and is equal to $\frac{1}{2\, \mathrm{Re}\left(z \right)} = \frac{1}{2r\cos \theta}$ assuming $z$ is not purely imaginary. If $z$ \textsl{is} purely imaginary, then $z$ must be equal to either $ir$ or $-ir$ (as $|z|=r$), in which case $z^2 + r^2=0$, and the given expression $\omega=\frac{z}{z^2 + r^2}$ is undefined.$

leehuan · Nov 28, 2015

InteGrand said:
$\textbf{ALTERNATE METHOD}$

$Since $\arg (z)$ is defined, we can assume $z\neq 0$. We will use the fact that a complex number $\omega \neq 0$ is real if and only if its reciprocal $\frac{1}{\omega}$ is real (this is easy to prove using polar form).$

$Denoting $\omega:= \frac{z}{z^2 + r^2}$, where $z=r\, \mathrm{cis}\left( \theta \right)$, we have$

$\begin{align*}\frac{1}{\omega} &= \frac{z^2 + r^2}{z} \\ &= z + \frac{r^2}{z} \\ &= z + \frac{r^2}{r \, \mathrm{cis}\left( \theta \right)} \\ &= z + \frac{r}{\mathrm{cis}\left( \theta \right)}\\ &= z + r \mathrm{cis} \left(- \theta \right) \quad (\text{since } \frac{1}{\mathrm{cis} (\theta)} = \left( \mathrm{cis} (\theta )\right) ^{-1} = \mathrm{cis} \left(- \theta \right) ) \\ &= z + \bar{z} \\ &= 2\, \mathrm{Re}\left(z\right) \end{align*}$

$which is real, so $\omega$ is also real, and is equal to $\frac{1}{2\, \mathrm{Re}\left(z \right)}$.$

Nice, I was wondering if reciprocating would be helpful but I was too scared to do it out of nowhere.

Ekman · Nov 28, 2015

Alternate Alternate Method:

$\frac{z}{z^2+r^2} = \frac{z(\bar{z}^2 + r^2)}{z^2 \bar{z}^2 + r^2(z^2+\bar{z}^2) + r^4} $ (multiplying top and bottom by the conjugate of the denominator) $$

$\frac{z(\bar{z}^2 + r^2)}{z^2 \bar{z}^2 + r^2(z^2+\bar{z}^2) + r^4} = \frac{r^2(z+\bar{z})}{2r^4 + r^2(z^2 + \bar{z}^2)} $ (Using: $ z\bar{z}=|z|^2 )$

$= \frac{2r^3 \cos{(\theta)}}{2r^4+ 2r^4 \cos{(2\theta)}} = \frac{\cos{(\theta)}}{r+r\cos{(2\theta)}} = \frac{1}{2r\cos{(\theta)}}$

$Hence Real$

Drsoccerball · Nov 28, 2015

Ekman said:
Alternate Alternate Method:

$\frac{z}{z^2+r^2} = \frac{z(\bar{z}^2 + r^2)}{z^2 \bar{z}^2 + r^2(z^2+\bar{z}^2) + r^4} $ (multiplying top and bottom by the conjugate of the denominator) $$

$\frac{z(\bar{z}^2 + r^2)}{z^2 \bar{z}^2 + r^2(z^2+\bar{z}^2) + r^4} = \frac{r^2(z+\bar{z})}{2r^4 + r^2(z^2 + \bar{z}^2)} $ (Using: $ z\bar{z}=|z|^2 )$

$= \frac{2r^3 \cos{(\theta)}}{2r^4+ 2r^4 \cos{(2\theta)}} = \frac{\cos{(\theta)}}{r+r\cos{(2\theta)}} = \frac{1}{2r\cos{(\theta)}}$

$Hence Real$

I thought of this method and was going to post it but ceebs ahah.

Make a Difference – Donate to Bored of Studies!

Stuck on Complex Numbers Question (1 Viewer)

Ambility

Active Member

leehuan

Well-Known Member

InteGrand

Well-Known Member

leehuan

Well-Known Member

Ekman

Well-Known Member

Drsoccerball

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)