Question

Complex Numbers

Answer 1

where is the question

Answer 2

Given that \[\Large w=\frac{z}{z^2+1}\]
and \[\Large z=x+iy\]
\[\Large \Im(z)=0\]

SHOW that \[\Large \left| z \right|=1\]

Answer 3

Is w omega ?

Answer 4

Yes

Answer 5

and what is J(z) = 0 ?

Answer 6

I think the question is show that \[\Large \left| \omega \right| =1\] actually.

also that means the imaginary part of z is equal to 0. It's an I

Answer 7

omega can be represented as an imaginary number.

Answer 8

Where would I go from there?

Answer 9

I've tried doing this:

\[\Large (z^2+1)\omega=z\]

Answer 10

Represent omega with its imaginary equivalent , then solve for x , y ( imaginary part = imaginary ) and (real = real)

next to get the norm of z sqrt(x^2 + y^2)

Answer 11

\[\omega = \frac{ 1 }{ 2 } \pm \frac{ \sqrt{3} }{ 2 }i\]

Answer 12

I'm not sure I follow.

Answer 13

w=a+bi for example,

I have to manipulate the RHS of the equation to get what I believe to just be x (since they say that the imaginary part is equal to 0) because when I take the magnitude of that, I will just get 1

Answer 14

And that is what the question wants from me

Answer 15

"I think the question is show that |ω|=1
actually.

also that means the imaginary part of z is equal to 0. It's an I"

if that's the case


|w| = sqrt( (0.5)^2 + ( sqrt(3)/2)^2) = sqrt( (0.25) + (0.75) = sqrt(1) = 1

Answer 16

Well I don't just get how you got those numbers

@hartnn any help?

Answer 17

imaginary part of z = 0
means y =0
right?

Answer 18

screenshot of the question?

Answer 19

z = x + 0i = x

and you want to prove that

|z| = 1 ?

x = 1 ?

and you induced that I(z) = 0 means that the imaginary part is zero.

Answer 20

@TrojanPoem I'm sorry. I wrote down the question wrong. The imaginary part of w is 0 not z; I know that manipulation will get me to that part where I will be left with x but yea I have to simplify the expression z/(z^2+1) first

Answer 21

Given that
w=z/(z^2+1)

and
z=x+iy

Im(w)=0


SHOW that
|z|=1

Answer 22

i have the long way,

Let w = u + 0i

wz^2 + w = z

now replace w with u and z with x+iy.

expand and compare real and imaginary parts.
shouldn't be difficult from here to show x^2+y^2 = 1 which actually implies |z| =1

Answer 23

I see the route of doing
\[\Large \omega=\frac{x+iy}{[(x+iy)^{2}+1]}\]

but that seems to be the longest possible way of achieving a result and this is just high school so I don't expect to solve it this way since my solutions never do the longest possible way.

Answer 24

x = u (1+x^2-y^2)
y = 2uxy

Answer 25

a shorter way would just use a short way to evaluate
Im (z/(z^2+1)) = 0
which won't be that short as compared to what we discussed

Answer 26

wz^2 + w = z

u (x+iy)^2 + u = x+iy

u(x^2-y^2 +2ixy) + u = x+iy

Comparing real and Imaginary parts.
x = u (1+x^2-y^2)
y = 2uxy

not so long afterall

Answer 27

I'm getting confused sorry.

I got to where you were,
x= u(1+x^2-y^2)
y=2uxy

x=u(1+x^2-y^2)
1=2ux

Answer 28

good eliminate u

u = 1/2x from 2nd equation
plug it in the first one

Answer 29

2x^2 = 1+x^2-y^2

x^2+y^2 =1
|z| =1

Answer 30

x^+y^2=1

Answer 31

ofcourse you'll have to take the square root on both sides for that

Answer 32

Oh yea, but then you plug in 0 for y^2 since the initial thing stated it equaled 0

Answer 33

nope
y or y^2 isn't 0.

Answer 34

Oh because sqrt(x^2+y^2) is the magnitude

Answer 35

yes
|z| = sqrt (x^2+y^2)