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anonymous
 one year ago
Complex Number Equation
anonymous
 one year ago
Complex Number Equation

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[z=x+yi\]\[z=1\]Prove that \(w\) is a real number, when \(w=\dfrac{z}{1+z^2}\) Note: \(y\ne0\) and \(1+z^2\ne0\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Let \(z=x+iy = re^{i\theta}\) then \[w=\dfrac{re^{i\theta}}{1+r^2e^{i2\theta}}\] multiply top and bottom by \(e^{i(\theta)}\) and get \[w = \dfrac{r}{e^{i(\theta)}+r^2e^{i\theta}}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1For \(w\) 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 : \[\sin(\theta) + r^2\sin(\theta) = 0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok, I have oficially confirmed that we haven't learnt the info needed to solve that question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure with \(e^{i\theta}\) etc means

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1you haven't learned polar form ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1ohk then forget about that, lets work it in rectangular form

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We have. Its given in \(z=r\times\text{cis}\times\theta\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But ok, I prefer rectangular

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Yes, \(re^{i\theta}\) is euler form for \(r*cis(\theta)\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Okay lets try and work it in rectangular

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1maybe start by plugging in \(z=x+iy\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[w=\dfrac{z}{1+z^2} = \dfrac{x+iy}{1+(x+iy)^2} = \dfrac{x+\color{red}{i}y}{1+x^2y^2 + 2xy\color{red}{i}}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1next multiply top and bottom by the conjugate of bottom

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1so that the denominator becomes real

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would we take the \(i\) as common, to make it easier?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1what is the conjugate of denominator ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\((x^2y^2+1)2xyi\)?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1yes multiply that top and bottom

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[\begin{align}w=\dfrac{z}{1+z^2} = \dfrac{x+iy}{1+(x+iy)^2} &= \dfrac{x+\color{red}{i}y}{1+x^2y^2 + 2xy\color{red}{i}}\\~\\ &=\dfrac{x+\color{red}{i}y}{1+x^2y^2 + 2xy\color{red}{i}}\times \dfrac{1+x^2y^2  2xy\color{red}{i}}{1+x^2y^2  2xy\color{red}{i}}\\~\\ &=\dfrac{x(1+x^2y^2+2xy^2) +\color{red}{i}y(1+x^2y^22x^2)}{(1+x^2y^2)^2+(2xy)^2} \end{align}\] set the imaginary part equal to 0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1because we're looking for a conditaion that makes w real

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1w is real means, its imaginary part is 0, yes ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So, wherever the is an imaginary, its equal to 0?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[\begin{align}w=\dfrac{z}{1+z^2} = \dfrac{x+iy}{1+(x+iy)^2} &= \dfrac{x+\color{red}{i}y}{1+x^2y^2 + 2xy\color{red}{i}}\\~\\ &=\dfrac{x+\color{red}{i}y}{1+x^2y^2 + 2xy\color{red}{i}}\times \dfrac{1+x^2y^2  2xy\color{red}{i}}{1+x^2y^2  2xy\color{red}{i}}\\~\\ &=\dfrac{x(1+x^2y^2+2xy^2) +\color{red}{i}\color{blue}{y(1+x^2y^22x^2)}}{\color{blue}{(1+x^2y^2)^2+(2xy)^2}} \end{align}\] thats the imaginary part, set that equal to 0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1*that blue thing is the imaginary part

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why is the bottom imaginary? I don't see an i

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1first write that in standard form

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1standard form of complex number : \(a+ib\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[\begin{align}w=\dfrac{z}{1+z^2} = \dfrac{x+iy}{1+(x+iy)^2} &= \dfrac{x+\color{red}{i}y}{1+x^2y^2 + 2xy\color{red}{i}}\\~\\ &=\dfrac{x+\color{red}{i}y}{1+x^2y^2 + 2xy\color{red}{i}}\times \dfrac{1+x^2y^2  2xy\color{red}{i}}{1+x^2y^2  2xy\color{red}{i}}\\~\\ &=\dfrac{x(1+x^2y^2+2xy^2) +\color{red}{i}\color{blue}{y(1+x^2y^22x^2)}}{\color{blue}{(1+x^2y^2)^2+(2xy)^2}}\\~\\ &=\dfrac{x(1+x^2y^2+2xy^2)}{\color{black}{(1+x^2y^2)^2+(2xy)^2}} +\color{red}{i}\dfrac{\color{blue}{y(1+x^2y^22x^2)}}{\color{blue}{(1+x^2y^2)^2+(2xy)^2}} \\~\\ \end{align}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\dfrac{x(1+x^2−y^2+2xy^2)}{(1+x^2−y^2)^2+(2xy)^2}+\dfrac{iy(1+x2−y2−2x2)}{(1+x^2−y^2)^2+(2xy)^2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Don't we just cancel the right fraction?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1we want to know when the imaginary part of \(w\) equals \(0\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok. That essentially just removes the right fraction, right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1idk what you mean by just removing the right fraction

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1we can't just remove right fraction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nevermind. Lets continue

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1set the imaginary part equal to 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok. So its just\[0=\dfrac{iy(1+x^2−y^2−2x^2) } {(1+x^2−y^2)^2+(2xy)^2}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1remove "i", imaginary part of w is \[\dfrac{y(1+x^2−y^2−2x^2) } {(1+x^2−y^2)^2+(2xy)^2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[{0=\frac{y(1+x^2−y^2−2x^2)}{(1+x^2−y^2)^2+(2xy)^2}}\]?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[0=y(1+x^2−y^2−2x^2)\]?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[0=y+x^2yy^32x^2y\] \[0=yx^2yy^3\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[0=y(1x^2y^2)\]\[0=1x^2y^2\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1from the hypothesis \(y\ne 0\) therefore \(x^2+y^2=1\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1which is same as \(z=1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And that's how we are meant to solve it? AWESOME :D

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Yes if we're allowed to use only rectangular form

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you SO MUCH @ganeshie8. I really do wonder how you solve like half of these questions ;)
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