Question

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.

Answer 1

@AravindG @amistre64 @ghazi @hartnn

Answer 2

oh too small latex ..my eye hurts

Answer 3

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

Answer 4

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 :)

Answer 5

Both conditions can be applied... @amistre64

Answer 6

how @AravindG ?

Answer 7

well I started like this :

\[\large{| \frac{\frac{z_1}{z_2} -1}{\frac{z_1}{z_2} +1}| = 1}\]

\[\large{\textbf{or} \space |\frac{z_1}{z_2} -1| = |\frac{z_1}{z_2}+1| }\]

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

oh god i was right :P

Answer 13

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

Answer 14

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

Answer 15

You can do it easily by interpretting is geometrically.

Answer 16

it*

Answer 17

im just not that attuned to complex calculations :)

Answer 18

|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 \(\dfrac{\pi}{2}\)

Answer 19

;) geometry is wonderful

Answer 20

wait...

Answer 21

can you draw that parallelogram please @AravindG

Answer 22

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

Answer 23

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

Answer 24

well right :) I got it ..

Answer 25

But what about the second one..

Answer 26

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

Answer 27

@mathslover where do you have confusion in my working ?

Answer 28

I am confused whether that parallelogram is possible or not?

Answer 29

can you draw that for me please?

Answer 30

just consider z1 and z2 as two vectors ....

P.S. I am bad at drawing

Answer 31

is it right diagram ?