Question

Question with Complex Numbers

Answer 1

If \(z = x + iy\) where \(y ≠ 0\) and \(1+ z^2 ≠ 0\), show that the number \(w=\dfrac{z}{1+z^2}\)is real only if \(|z| = 1\)

Answer 2

would it help if i prove the converse?

Answer 3

What is the converse, @divu.mkr?

Answer 4

it's the reversed version of a statement

Answer 5

that if |z| = 1 then 'w' is real?

Answer 6

for example. if A then B is the original statement
the converse would be If B then A
and I'm going in and out of the question, my openstudy is real bad

Answer 7

Sure, if that works @divu.mkr

Answer 8

I dont know how to use latex... so try to understand
just use this (z)(z bar) = |z|^2 so z=1/zbar
and if a complex number is real then w= w bar
take bar on both the sides replace 'z bar' by z
they will become same

Answer 9

So, \(z\times \overline{z}=|z|^2\)
Therefore, \(z=\dfrac{1}{\overline{z}}\)

I don't get what you mean if a complex number is real, then w = w bar

Answer 10

when we take a bar over a complex number the sign of the complex quantity reverses
so if the complex number is purely real then there will be no change in the complex number as the complex quantity is absent

Answer 11

\[\bar{z} \]

Answer 12

latex is \bar{insertsomethinghere} for the bar over the area

Answer 13

Ahh, I see. So, \(w\) is a complex number (cause you are dividing a complex by a complex), but since we have established its real, its conjugate should be the same..?

Answer 14

Cause there is no imaginary component?

Answer 15

yep, exactly

Answer 16

Great :)
So, we turn:
\[\huge{w=\frac{z}{1+z^2}}\]
To
\[\huge{\overline{w}=\frac{\overline{z}}{\overline{1+z^2}}}\]

Answer 17

Right..?

Answer 18

yep now just one of them

Answer 19

\[\huge{w}=\frac{\frac{1}{z}}{1+(\frac{1}{z})^2}\]?

Answer 20

yeah doing good

Answer 21

but it will be w bar

Answer 22

Didn'w we say w = w bar tho?

Answer 23

@divu.mkr?

Answer 24

you substituted in wrong eqn ...just saying that