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oh too small latex ..my eye hurts

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

Both conditions can be applied... @amistre64

how @AravindG ?

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

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

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

oh god i was right :P

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

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

You can do it easily by interpretting is geometrically.

it*

im just not that attuned to complex calculations :)

;) geometry is wonderful

wait...

can you draw that parallelogram please @AravindG

well right :) I got it ..

But what about the second one..

@mathslover where do you have confusion in my working ?

I am confused whether that parallelogram is possible or not?

can you draw that for me please?

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

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

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

gosh u made me draw that ...:P

|dw:1361019850236:dw|

is C (z_1 - z_2)

c is z1+z2

oh got it now :)

:)