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 one year 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.
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.2@AravindG @amistre64 @ghazi @hartnn

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

amistre64
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.2Both conditions can be applied... @amistre64

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

amistre64
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1i think I made a mistake there ..wait lemme think

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

AravindG
 one year ago
Best ResponseYou've already chosen the best response.1oh god i was right :P

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

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

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

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

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

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

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

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

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

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

AravindG
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.2dw:1361019499183:dw is it right diagram ?

mathslover
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1i would say this : dw:1361019666403:dw

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

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