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mathslover
 3 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.
mathslover
 3 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.

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mathslover
 3 years ago
Best ResponseYou've already chosen the best response.2@AravindG @amistre64 @ghazi @hartnn

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.1oh too small latex ..my eye hurts

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0isnt a general complex number: a + bi ? or do we need to define it in trig terms?

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.2If \(\large{z_1}\) 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 :)

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.2Both conditions can be applied... @amistre64

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.2well I started like this : \[\large{ \frac{\frac{z_1}{z_2} 1}{\frac{z_1}{z_2} +1} = 1}\] \[\large{\textbf{or} \space \frac{z_1}{z_2} 1 = \frac{z_1}{z_2}+1 }\]

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.2I am thinking of squaring both sides, should I go for it ?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{(a+bi)(n+mi)}{(a+bi)+(n+mi)}=1\] \[(a+bi)(n+mi)=(a+bi)+(n+mi)\] \[(n+mi)=(n+mi)\] \[0=2(n+mi)\] just a thought ....

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.1i think I made a mistake there ..wait lemme think

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.2But amistre where did modulus go ? In LHS ...

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.2\[\large{\frac{z_1z_2}{z_1+z_2} = 1 \ne \frac{z_1 z_2}{z_1+z_2} =1 }\]

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.2^ not necessary that : (a+bi)(n+mi) / (a+bi)+(n+mi) = 1 ...

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0You can do it easily by interpretting is geometrically.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0im just not that attuned to complex calculations :)

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.1z1+z2 =z1z2 this means lenth of the vectors z1+z2 and z1z2 are equal Now think of z1 and z2 as two vectors z1+z2 is one diagonal of the parallelogram ,z1z2 is the other diagonal Given the diagonal lengths are equal thus the parallelogram is a rectangle So angle between z1 and z2 is \(\dfrac{\pi}{2}\)

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.1;) geometry is wonderful

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.2can you draw that parallelogram please @AravindG

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.2I am confused that whether that parallelogram is possible or not with two diagonals as : z1 + z2 and z1z2

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1\(\huge \frac{(a+bi)(n+mi)}{(a+bi)+(n+mi)}=1 \\ (an)^2+(bm)^2=(a+n)^2+(b+n)^2 \\ \implies an+bm=0\) \(\huge \frac{i(a+bi)}{(n+mi)}=\frac{i(a+bi)}{(n+mi)}\dfrac{(nmi)}{(nmi)}= \\ \huge =\dfrac{i(an+bminb+ami)}{(n^2+m^2)}=\dfrac{i^2(0+am+nb)}{n^2+m^2}= \\ =k\)

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.2well right :) I got it ..

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.2But what about the second one..

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.2I 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

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.1@mathslover where do you have confusion in my working ?

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.2I am confused whether that parallelogram is possible or not?

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.2can you draw that for me please?

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.1just consider z1 and z2 as two vectors .... P.S. I am bad at drawing

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1361019499183:dw is it right diagram ?

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.2instead 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)) ...

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.1i would say this : dw:1361019666403:dw

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.1gosh u made me draw that ...:P

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1361019850236:dw
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