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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
 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
 one year ago

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

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

amistre64Best ResponseYou've already chosen the best response.0
isnt a general complex number: a + bi ? or do we need to define it in trig terms?
 one year ago

mathsloverBest ResponseYou've already chosen the best response.2
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 :)
 one year ago

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

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

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

amistre64Best 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 ....
 one year ago

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

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

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

mathsloverBest ResponseYou've already chosen the best response.2
\[\large{\frac{z_1z_2}{z_1+z_2} = 1 \ne \frac{z_1 z_2}{z_1+z_2} =1 }\]
 one year ago

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

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

amistre64Best ResponseYou've already chosen the best response.0
im just not that attuned to complex calculations :)
 one year ago

AravindGBest ResponseYou've already chosen the best response.1
z1+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 \(\dfrac{\pi}{2}\)
 one year ago

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

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

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

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

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

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

mathsloverBest ResponseYou've already chosen the best response.2
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
 one year ago

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

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

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

AravindGBest ResponseYou've already chosen the best response.1
just consider z1 and z2 as two vectors .... P.S. I am bad at drawing
 one year ago

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

mathsloverBest ResponseYou've already chosen the best response.2
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)) ...
 one year ago

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

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

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