Ahsome
  • Ahsome
Complex Number Equation
Mathematics
katieb
  • katieb
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Ahsome
  • Ahsome
\[z=x+yi\]\[|z|=1\]Prove that \(w\) is a real number, when \(w=\dfrac{z}{1+z^2}\) Note: \(y\ne0\) and \(1+z^2\ne0\)
Ahsome
  • Ahsome
ganeshie8
  • ganeshie8
Let \(z=x+iy = re^{i\theta}\) then \[w=\dfrac{re^{i\theta}}{1+r^2e^{i2\theta}}\] multiply top and bottom by \(e^{i(-\theta)}\) and get \[w = \dfrac{r}{e^{i(-\theta)}+r^2e^{i\theta}}\]

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ganeshie8
  • ganeshie8
For \(w\) to be a real number, it must be the case that the denominator is real, so set the imaginary part of denominator equal to 0 : \[\sin(-\theta) + r^2\sin(\theta) = 0\]
Ahsome
  • Ahsome
Ok, I have oficially confirmed that we haven't learnt the info needed to solve that question
Ahsome
  • Ahsome
I'm not sure with \(e^{i\theta}\) etc means
ganeshie8
  • ganeshie8
you haven't learned polar form ?
ganeshie8
  • ganeshie8
ohk then forget about that, lets work it in rectangular form
Ahsome
  • Ahsome
We have. Its given in \(z=r\times\text{cis}\times\theta\)
Ahsome
  • Ahsome
But ok, I prefer rectangular
ganeshie8
  • ganeshie8
Yes, \(re^{i\theta}\) is euler form for \(r*cis(\theta)\)
ganeshie8
  • ganeshie8
Okay lets try and work it in rectangular
Ahsome
  • Ahsome
Ok then :)
ganeshie8
  • ganeshie8
maybe start by plugging in \(z=x+iy\)
Ahsome
  • Ahsome
Into the fraction?
ganeshie8
  • ganeshie8
\[w=\dfrac{z}{1+z^2} = \dfrac{x+iy}{1+(x+iy)^2} = \dfrac{x+\color{red}{i}y}{1+x^2-y^2 + 2xy\color{red}{i}}\]
ganeshie8
  • ganeshie8
next multiply top and bottom by the conjugate of bottom
ganeshie8
  • ganeshie8
so that the denominator becomes real
Ahsome
  • Ahsome
Would we take the \(i\) as common, to make it easier?
ganeshie8
  • ganeshie8
what is the conjugate of denominator ?
Ahsome
  • Ahsome
\((x^2-y^2+1)-2xyi\)?
ganeshie8
  • ganeshie8
yes multiply that top and bottom
ganeshie8
  • ganeshie8
\[\begin{align}w=\dfrac{z}{1+z^2} = \dfrac{x+iy}{1+(x+iy)^2} &= \dfrac{x+\color{red}{i}y}{1+x^2-y^2 + 2xy\color{red}{i}}\\~\\ &=\dfrac{x+\color{red}{i}y}{1+x^2-y^2 + 2xy\color{red}{i}}\times \dfrac{1+x^2-y^2 - 2xy\color{red}{i}}{1+x^2-y^2 - 2xy\color{red}{i}}\\~\\ &=\dfrac{x(1+x^2-y^2+2xy^2) +\color{red}{i}y(1+x^2-y^2-2x^2)}{(1+x^2-y^2)^2+(2xy)^2} \end{align}\] set the imaginary part equal to 0
Ahsome
  • Ahsome
Why?
ganeshie8
  • ganeshie8
because we're looking for a conditaion that makes w real
ganeshie8
  • ganeshie8
w is real means, its imaginary part is 0, yes ?
Ahsome
  • Ahsome
Ahh, ok
Ahsome
  • Ahsome
So, wherever the is an imaginary, its equal to 0?
ganeshie8
  • ganeshie8
\[\begin{align}w=\dfrac{z}{1+z^2} = \dfrac{x+iy}{1+(x+iy)^2} &= \dfrac{x+\color{red}{i}y}{1+x^2-y^2 + 2xy\color{red}{i}}\\~\\ &=\dfrac{x+\color{red}{i}y}{1+x^2-y^2 + 2xy\color{red}{i}}\times \dfrac{1+x^2-y^2 - 2xy\color{red}{i}}{1+x^2-y^2 - 2xy\color{red}{i}}\\~\\ &=\dfrac{x(1+x^2-y^2+2xy^2) +\color{red}{i}\color{blue}{y(1+x^2-y^2-2x^2)}}{\color{blue}{(1+x^2-y^2)^2+(2xy)^2}} \end{align}\] thats the imaginary part, set that equal to 0
ganeshie8
  • ganeshie8
*that blue thing is the imaginary part
Ahsome
  • Ahsome
Why is the bottom imaginary? I don't see an i
ganeshie8
  • ganeshie8
first write that in standard form
ganeshie8
  • ganeshie8
standard form of complex number : \(a+ib\)
Ahsome
  • Ahsome
Yup
ganeshie8
  • ganeshie8
\[\begin{align}w=\dfrac{z}{1+z^2} = \dfrac{x+iy}{1+(x+iy)^2} &= \dfrac{x+\color{red}{i}y}{1+x^2-y^2 + 2xy\color{red}{i}}\\~\\ &=\dfrac{x+\color{red}{i}y}{1+x^2-y^2 + 2xy\color{red}{i}}\times \dfrac{1+x^2-y^2 - 2xy\color{red}{i}}{1+x^2-y^2 - 2xy\color{red}{i}}\\~\\ &=\dfrac{x(1+x^2-y^2+2xy^2) +\color{red}{i}\color{blue}{y(1+x^2-y^2-2x^2)}}{\color{blue}{(1+x^2-y^2)^2+(2xy)^2}}\\~\\ &=\dfrac{x(1+x^2-y^2+2xy^2)}{\color{black}{(1+x^2-y^2)^2+(2xy)^2}} +\color{red}{i}\dfrac{\color{blue}{y(1+x^2-y^2-2x^2)}}{\color{blue}{(1+x^2-y^2)^2+(2xy)^2}} \\~\\ \end{align}\]
Ahsome
  • Ahsome
\[\dfrac{x(1+x^2−y^2+2xy^2)}{(1+x^2−y^2)^2+(2xy)^2}+\dfrac{iy(1+x2−y2−2x2)}{(1+x^2−y^2)^2+(2xy)^2}\]
Ahsome
  • Ahsome
Don't we just cancel the right fraction?
ganeshie8
  • ganeshie8
set that equal to 0
ganeshie8
  • ganeshie8
we want to know when the imaginary part of \(w\) equals \(0\)
Ahsome
  • Ahsome
Ok. That essentially just removes the right fraction, right?
ganeshie8
  • ganeshie8
idk what you mean by just removing the right fraction
ganeshie8
  • ganeshie8
we can't just remove right fraction
Ahsome
  • Ahsome
Nevermind. Lets continue
ganeshie8
  • ganeshie8
set the imaginary part equal to 0
Ahsome
  • Ahsome
Ok. So its just\[0=\dfrac{iy(1+x^2−y^2−2x^2) } {(1+x^2−y^2)^2+(2xy)^2}\]
ganeshie8
  • ganeshie8
remove "i", imaginary part of w is \[\dfrac{y(1+x^2−y^2−2x^2) } {(1+x^2−y^2)^2+(2xy)^2}\]
ganeshie8
  • ganeshie8
set that equal to 0
Ahsome
  • Ahsome
\[{0=\frac{y(1+x^2−y^2−2x^2)}{(1+x^2−y^2)^2+(2xy)^2}}\]?
ganeshie8
  • ganeshie8
Yep
Ahsome
  • Ahsome
\[0=y(1+x^2−y^2−2x^2)\]?
ganeshie8
  • ganeshie8
Yes
Ahsome
  • Ahsome
\[0=y+x^2y-y^3-2x^2y\] \[0=y-x^2y-y^3\]
ganeshie8
  • ganeshie8
factor out y
Ahsome
  • Ahsome
\[0=y(1-x^2-y^2)\]\[0=1-x^2-y^2\]
Ahsome
  • Ahsome
\[x^2+y^2=1\]
ganeshie8
  • ganeshie8
Looks good!
ganeshie8
  • ganeshie8
from the hypothesis \(y\ne 0\) therefore \(x^2+y^2=1\)
ganeshie8
  • ganeshie8
which is same as \(|z|=1\)
Ahsome
  • Ahsome
And that's how we are meant to solve it? AWESOME :D
ganeshie8
  • ganeshie8
Yes if we're allowed to use only rectangular form
Ahsome
  • Ahsome
Great
Ahsome
  • Ahsome
Thank you SO MUCH @ganeshie8. I really do wonder how you solve like half of these questions ;)
ganeshie8
  • ganeshie8
np:)

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