Question

Complex Number Equation

Answer 1

\[z=x+yi\]\[|z|=1\]Prove that \(w\) is a real number, when \(w=\dfrac{z}{1+z^2}\)

Note: \(y\ne0\) and \(1+z^2\ne0\)

Answer 2

@ganeshie8

Answer 3

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
\[w = \dfrac{r}{e^{i(-\theta)}+r^2e^{i\theta}}\]

Answer 4

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 :
\[\sin(-\theta) + r^2\sin(\theta) = 0\]

Answer 5

Ok, I have oficially confirmed that we haven't learnt the info needed to solve that question

Answer 6

I'm not sure with \(e^{i\theta}\) etc means

Answer 7

you haven't learned polar form ?

Answer 8

ohk then forget about that, lets work it in rectangular form

Answer 9

We have. Its given in \(z=r\times\text{cis}\times\theta\)

Answer 10

But ok, I prefer rectangular

Answer 11

Yes, \(re^{i\theta}\) is euler form for \(r*cis(\theta)\)

Answer 12

Okay lets try and work it in rectangular

Answer 13

Ok then :)

Answer 14

maybe start by plugging in \(z=x+iy\)

Answer 15

Into the fraction?

Answer 16

\[w=\dfrac{z}{1+z^2} = \dfrac{x+iy}{1+(x+iy)^2} = \dfrac{x+\color{red}{i}y}{1+x^2-y^2 + 2xy\color{red}{i}}\]

Answer 17

next multiply top and bottom by the conjugate of bottom

Answer 18

so that the denominator becomes real

Answer 19

Would we take the \(i\) as common, to make it easier?

Answer 20

what is the conjugate of denominator ?

Answer 21

\((x^2-y^2+1)-2xyi\)?

Answer 22

yes multiply that top and bottom

Answer 23

\[\begin{align}w=\dfrac{z}{1+z^2} = \dfrac{x+iy}{1+(x+iy)^2} &= \dfrac{x+\color{red}{i}y}{1+x^2-y^2 + 2xy\color{red}{i}}\\~\\
&=\dfrac{x+\color{red}{i}y}{1+x^2-y^2 + 2xy\color{red}{i}}\times \dfrac{1+x^2-y^2 - 2xy\color{red}{i}}{1+x^2-y^2 - 2xy\color{red}{i}}\\~\\
&=\dfrac{x(1+x^2-y^2+2xy^2) +\color{red}{i}y(1+x^2-y^2-2x^2)}{(1+x^2-y^2)^2+(2xy)^2}
\end{align}\]

set the imaginary part equal to 0

Answer 24

Why?

Answer 25

because we're looking for a conditaion that makes w real

Answer 26

w is real means, its imaginary part is 0, yes ?

Answer 27

Ahh, ok

Answer 28

So, wherever the is an imaginary, its equal to 0?

Answer 29

\[\begin{align}w=\dfrac{z}{1+z^2} = \dfrac{x+iy}{1+(x+iy)^2} &= \dfrac{x+\color{red}{i}y}{1+x^2-y^2 + 2xy\color{red}{i}}\\~\\
&=\dfrac{x+\color{red}{i}y}{1+x^2-y^2 + 2xy\color{red}{i}}\times \dfrac{1+x^2-y^2 - 2xy\color{red}{i}}{1+x^2-y^2 - 2xy\color{red}{i}}\\~\\
&=\dfrac{x(1+x^2-y^2+2xy^2) +\color{red}{i}\color{blue}{y(1+x^2-y^2-2x^2)}}{\color{blue}{(1+x^2-y^2)^2+(2xy)^2}}
\end{align}\]

thats the imaginary part, set that equal to 0

Answer 30

*that blue thing is the imaginary part

Answer 31

Why is the bottom imaginary? I don't see an i

Answer 32

first write that in standard form

Answer 33

standard form of complex number : \(a+ib\)

Answer 34

Yup

Answer 35

\[\begin{align}w=\dfrac{z}{1+z^2} = \dfrac{x+iy}{1+(x+iy)^2} &= \dfrac{x+\color{red}{i}y}{1+x^2-y^2 + 2xy\color{red}{i}}\\~\\
&=\dfrac{x+\color{red}{i}y}{1+x^2-y^2 + 2xy\color{red}{i}}\times \dfrac{1+x^2-y^2 - 2xy\color{red}{i}}{1+x^2-y^2 - 2xy\color{red}{i}}\\~\\
&=\dfrac{x(1+x^2-y^2+2xy^2) +\color{red}{i}\color{blue}{y(1+x^2-y^2-2x^2)}}{\color{blue}{(1+x^2-y^2)^2+(2xy)^2}}\\~\\
&=\dfrac{x(1+x^2-y^2+2xy^2)}{\color{black}{(1+x^2-y^2)^2+(2xy)^2}} +\color{red}{i}\dfrac{\color{blue}{y(1+x^2-y^2-2x^2)}}{\color{blue}{(1+x^2-y^2)^2+(2xy)^2}} \\~\\
\end{align}\]

Answer 36

\[\dfrac{x(1+x^2−y^2+2xy^2)}{(1+x^2−y^2)^2+(2xy)^2}+\dfrac{iy(1+x2−y2−2x2)}{(1+x^2−y^2)^2+(2xy)^2}\]

Answer 37

Don't we just cancel the right fraction?

Answer 38

set that equal to 0

Answer 39

we want to know when the imaginary part of \(w\) equals \(0\)

Answer 40

Ok. That essentially just removes the right fraction, right?

Answer 41

idk what you mean by just removing the right fraction

Answer 42

we can't just remove right fraction

Answer 43

Nevermind. Lets continue

Answer 44

set the imaginary part equal to 0

Answer 45

Ok. So its just\[0=\dfrac{iy(1+x^2−y^2−2x^2)
}
{(1+x^2−y^2)^2+(2xy)^2}\]

Answer 46

remove "i", imaginary part of w is

\[\dfrac{y(1+x^2−y^2−2x^2)
}
{(1+x^2−y^2)^2+(2xy)^2}\]

Answer 47

set that equal to 0

Answer 48

\[{0=\frac{y(1+x^2−y^2−2x^2)}{(1+x^2−y^2)^2+(2xy)^2}}\]?

Answer 49

Yep

Answer 50

\[0=y(1+x^2−y^2−2x^2)\]?

Answer 51

Yes

Answer 52

\[0=y+x^2y-y^3-2x^2y\]
\[0=y-x^2y-y^3\]

Answer 53

factor out y

Answer 54

\[0=y(1-x^2-y^2)\]\[0=1-x^2-y^2\]

Answer 55

\[x^2+y^2=1\]

Answer 56

Looks good!

Answer 57

from the hypothesis \(y\ne 0\)
therefore \(x^2+y^2=1\)

Answer 58

which is same as \(|z|=1\)

Answer 59

And that's how we are meant to solve it?
AWESOME :D

Answer 60

Yes if we're allowed to use only rectangular form

Answer 61

Great

Answer 62

Thank you SO MUCH @ganeshie8. I really do wonder how you solve like half of these questions ;)

Answer 63

np:)