Ahsome
  • Ahsome
Question with Complex Numbers
Mathematics
schrodinger
  • schrodinger
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Ahsome
  • Ahsome
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 \(|z| = 1\)
anonymous
  • anonymous
would it help if i prove the converse?
Ahsome
  • Ahsome
What is the converse, @divu.mkr?

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UsukiDoll
  • UsukiDoll
it's the reversed version of a statement
anonymous
  • anonymous
that if |z| = 1 then 'w' is real?
UsukiDoll
  • UsukiDoll
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
Ahsome
  • Ahsome
Sure, if that works @divu.mkr
anonymous
  • anonymous
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
Ahsome
  • Ahsome
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
anonymous
  • anonymous
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
UsukiDoll
  • UsukiDoll
\[\bar{z} \]
UsukiDoll
  • UsukiDoll
latex is \bar{insertsomethinghere} for the bar over the area
Ahsome
  • Ahsome
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..?
Ahsome
  • Ahsome
Cause there is no imaginary component?
anonymous
  • anonymous
yep, exactly
Ahsome
  • Ahsome
Great :) So, we turn: \[\huge{w=\frac{z}{1+z^2}}\] To \[\huge{\overline{w}=\frac{\overline{z}}{\overline{1+z^2}}}\]
Ahsome
  • Ahsome
Right..?
anonymous
  • anonymous
yep now just one of them
Ahsome
  • Ahsome
\[\huge{w}=\frac{\frac{1}{z}}{1+(\frac{1}{z})^2}\]?
anonymous
  • anonymous
yeah doing good
anonymous
  • anonymous
but it will be w bar
Ahsome
  • Ahsome
Didn'w we say w = w bar tho?
Ahsome
  • Ahsome
@divu.mkr?
anonymous
  • anonymous
you substituted in wrong eqn ...just saying that

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