mathslover 2 years ago If $$z_1$$ and $$z_2$$ are two complex numbers such that $$|\frac{z_1 - z_2}{z_1 + z_2}| =1$$ , prove that, $$\frac{iz_1}{z_2}=k$$ is a real number. Find the angle between the lines from the origin to the points $$z_1 + z_2$$ and $$z_1 - z_2$$ in terms of k.

1. mathslover

@AravindG @amistre64 @ghazi @hartnn

2. AravindG

oh too small latex ..my eye hurts

3. amistre64

isnt a general complex number: a + bi ? or do we need to define it in trig terms?

4. mathslover

If $$\large{z_1}$$ and $$\large{z_2}$$ are two complex numbers such that $$\large{|\frac{z_1 - z_2}{z_1 + z_2}| = 1 }$$ , prove that , $$\large{\frac{i z_1}{z_2} = k }$$ , where k is a real number. Find the angle between the lines from the origin to the points $$\large{z_1 + z_2}$$ and $$\large{z_1 - z_2}$$ in terms of k... It did hurt me too :)

5. mathslover

Both conditions can be applied... @amistre64

6. mathslover

how @AravindG ?

7. mathslover

well I started like this : $\large{| \frac{\frac{z_1}{z_2} -1}{\frac{z_1}{z_2} +1}| = 1}$ $\large{\textbf{or} \space |\frac{z_1}{z_2} -1| = |\frac{z_1}{z_2}+1| }$

8. mathslover

I am thinking of squaring both sides, should I go for it ?

9. amistre64

$\frac{(a+bi)-(n+mi)}{(a+bi)+(n+mi)}=1$ $(a+bi)-(n+mi)=(a+bi)+(n+mi)$ $-(n+mi)=(n+mi)$ $0=2(n+mi)$ just a thought ....

10. AravindG

i think I made a mistake there ..wait lemme think

11. mathslover

But amistre where did modulus go ? In LHS ...

12. AravindG

oh god i was right :P

13. mathslover

$\large{|\frac{z_1-z_2}{z_1+z_2}| = 1 \ne \frac{z_1 -z_2}{z_1+z_2} =1 }$

14. mathslover

^ not necessary that : (a+bi)-(n+mi) / (a+bi)+(n+mi) = 1 ...

15. shubhamsrg

You can do it easily by interpretting is geometrically.

16. shubhamsrg

it*

17. amistre64

im just not that attuned to complex calculations :)

18. AravindG

|z1+z2| =|z1-z2| this means lenth of the vectors z1+z2 and z1-z2 are equal Now think of z1 and z2 as two vectors z1+z2 is one diagonal of the parallelogram ,z1-z2 is the other diagonal Given the diagonal lengths are equal thus the parallelogram is a rectangle So angle between z1 and z2 is $$\dfrac{\pi}{2}$$

19. AravindG

;) geometry is wonderful

20. mathslover

wait...

21. mathslover

can you draw that parallelogram please @AravindG

22. mathslover

I am confused that whether that parallelogram is possible or not with two diagonals as : z1 + z2 and z1-z2

23. hartnn

$$\huge |\frac{(a+bi)-(n+mi)}{(a+bi)+(n+mi)}|=1 \\ (a-n)^2+(b-m)^2=(a+n)^2+(b+n)^2 \\ \implies an+bm=0$$ $$\huge \frac{i(a+bi)}{(n+mi)}=\frac{i(a+bi)}{(n+mi)}\dfrac{(n-mi)}{(n-mi)}= \\ \huge =\dfrac{i(an+bm-inb+ami)}{(n^2+m^2)}=\dfrac{i^2(0+am+nb)}{n^2+m^2}= \\ =k$$

24. mathslover

well right :) I got it ..

25. mathslover

But what about the second one..

26. mathslover

I did the first one like this : $\large{|\frac{z_1}{z_2} - 1| = | \frac{z_1}{z_2} + 1|}$ Squaring both sides : $\large{|\frac{z_1}{z_2}|^2 + 1 - 2Re (\frac{z_1}{z_2}) = |\frac{z_1}{z_2}|^2 + 1 + 2 Re(\frac{z_1}{z_2})}$ $\large{4Re(\frac{z_1}{z_2}) = 0}$ $\large{Re(\frac{z_1}{z_2}) = 0}$ therefore (z_1)/z_2 is purely imaginary number. Therefore $\large{\frac{z_1}{z_2} = i \frac{z_1}{z_2} = k}$ where k is a real number

27. AravindG

@mathslover where do you have confusion in my working ?

28. mathslover

I am confused whether that parallelogram is possible or not?

29. mathslover

can you draw that for me please?

30. AravindG

just consider z1 and z2 as two vectors .... P.S. I am bad at drawing

31. mathslover

|dw:1361019499183:dw| is it right diagram ?

32. mathslover

instead of drawing can you tell me that what are co ordinates of the parallelogram : like A(z_1) B(z_2) C(z_(-1)) D(z_(-2)) ...

33. AravindG

i would say this : |dw:1361019666403:dw|

34. AravindG

gosh u made me draw that ...:P

35. mathslover

|dw:1361019850236:dw|

36. mathslover

is C (z_1 - z_2)

37. AravindG

c is z1+z2

38. mathslover

oh got it now :)

39. AravindG

:)