## Ahsome one year ago Question with Complex Numbers

1. 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 $$|z| = 1$$

2. anonymous

would it help if i prove the converse?

3. ahsome

What is the converse, @divu.mkr?

4. UsukiDoll

it's the reversed version of a statement

5. anonymous

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

6. 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

7. ahsome

Sure, if that works @divu.mkr

8. 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

9. 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

10. 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

11. UsukiDoll

$\bar{z}$

12. UsukiDoll

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

13. 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..?

14. ahsome

Cause there is no imaginary component?

15. anonymous

yep, exactly

16. ahsome

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

17. ahsome

Right..?

18. anonymous

yep now just one of them

19. ahsome

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

20. anonymous

yeah doing good

21. anonymous

but it will be w bar

22. ahsome

Didn'w we say w = w bar tho?

23. ahsome

@divu.mkr?

24. anonymous

you substituted in wrong eqn ...just saying that