## mathslover Group Title 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. one year ago one year ago

1. mathslover Group Title

@AravindG @amistre64 @ghazi @hartnn

2. AravindG Group Title

oh too small latex ..my eye hurts

3. amistre64 Group Title

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

4. mathslover Group Title

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 Group Title

Both conditions can be applied... @amistre64

6. mathslover Group Title

how @AravindG ?

7. mathslover Group Title

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 Group Title

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

9. amistre64 Group Title

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

10. AravindG Group Title

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

11. mathslover Group Title

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

12. AravindG Group Title

oh god i was right :P

13. mathslover Group Title

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

14. mathslover Group Title

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

15. shubhamsrg Group Title

You can do it easily by interpretting is geometrically.

16. shubhamsrg Group Title

it*

17. amistre64 Group Title

im just not that attuned to complex calculations :)

18. AravindG Group Title

|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 Group Title

;) geometry is wonderful

20. mathslover Group Title

wait...

21. mathslover Group Title

can you draw that parallelogram please @AravindG

22. mathslover Group Title

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

23. hartnn Group Title

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

24. mathslover Group Title

well right :) I got it ..

25. mathslover Group Title

But what about the second one..

26. mathslover Group Title

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 Group Title

@mathslover where do you have confusion in my working ?

28. mathslover Group Title

I am confused whether that parallelogram is possible or not?

29. mathslover Group Title

can you draw that for me please?

30. AravindG Group Title

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

31. mathslover Group Title

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

32. mathslover Group Title

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 Group Title

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

34. AravindG Group Title

gosh u made me draw that ...:P

35. mathslover Group Title

|dw:1361019850236:dw|

36. mathslover Group Title

is C (z_1 - z_2)

37. AravindG Group Title

c is z1+z2

38. mathslover Group Title

oh got it now :)

39. AravindG Group Title

:)