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mathslover
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.
@AravindG @amistre64 @ghazi @hartnn
oh too small latex ..my eye hurts
isnt a general complex number: a + bi ? or do we need to define it in trig terms?
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 :)
Both conditions can be applied... @amistre64
well I started like this : \[\large{| \frac{\frac{z_1}{z_2} -1}{\frac{z_1}{z_2} +1}| = 1}\] \[\large{\textbf{or} \space |\frac{z_1}{z_2} -1| = |\frac{z_1}{z_2}+1| }\]
I am thinking of squaring both sides, should I go for it ?
\[\frac{(a+bi)-(n+mi)}{(a+bi)+(n+mi)}=1\] \[(a+bi)-(n+mi)=(a+bi)+(n+mi)\] \[-(n+mi)=(n+mi)\] \[0=2(n+mi)\] just a thought ....
i think I made a mistake there ..wait lemme think
But amistre where did modulus go ? In LHS ...
\[\large{|\frac{z_1-z_2}{z_1+z_2}| = 1 \ne \frac{z_1 -z_2}{z_1+z_2} =1 }\]
^ not necessary that : (a+bi)-(n+mi) / (a+bi)+(n+mi) = 1 ...
You can do it easily by interpretting is geometrically.
im just not that attuned to complex calculations :)
|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 \(\dfrac{\pi}{2}\)
;) geometry is wonderful
can you draw that parallelogram please @AravindG
I am confused that whether that parallelogram is possible or not with two diagonals as : z1 + z2 and z1-z2
\(\huge |\frac{(a+bi)-(n+mi)}{(a+bi)+(n+mi)}|=1 \\ (a-n)^2+(b-m)^2=(a+n)^2+(b+n)^2 \\ \implies an+bm=0\) \(\huge \frac{i(a+bi)}{(n+mi)}=\frac{i(a+bi)}{(n+mi)}\dfrac{(n-mi)}{(n-mi)}= \\ \huge =\dfrac{i(an+bm-inb+ami)}{(n^2+m^2)}=\dfrac{i^2(0+am+nb)}{n^2+m^2}= \\ =k\)
well right :) I got it ..
But what about the second one..
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
@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 ?
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)) ...
i would say this : |dw:1361019666403:dw|
gosh u made me draw that ...:P
|dw:1361019850236:dw|