mathslover
  • 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.
Mathematics
schrodinger
  • schrodinger
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

mathslover
  • mathslover
@AravindG @amistre64 @ghazi @hartnn
AravindG
  • AravindG
oh too small latex ..my eye hurts
amistre64
  • amistre64
isnt a general complex number: a + bi ? or do we need to define it in trig terms?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

mathslover
  • mathslover
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 :)
mathslover
  • mathslover
Both conditions can be applied... @amistre64
mathslover
  • mathslover
how @AravindG ?
mathslover
  • mathslover
well I started like this : \[\large{| \frac{\frac{z_1}{z_2} -1}{\frac{z_1}{z_2} +1}| = 1}\] \[\large{\textbf{or} \space |\frac{z_1}{z_2} -1| = |\frac{z_1}{z_2}+1| }\]
mathslover
  • mathslover
I am thinking of squaring both sides, should I go for it ?
amistre64
  • amistre64
\[\frac{(a+bi)-(n+mi)}{(a+bi)+(n+mi)}=1\] \[(a+bi)-(n+mi)=(a+bi)+(n+mi)\] \[-(n+mi)=(n+mi)\] \[0=2(n+mi)\] just a thought ....
AravindG
  • AravindG
i think I made a mistake there ..wait lemme think
mathslover
  • mathslover
But amistre where did modulus go ? In LHS ...
AravindG
  • AravindG
oh god i was right :P
mathslover
  • mathslover
\[\large{|\frac{z_1-z_2}{z_1+z_2}| = 1 \ne \frac{z_1 -z_2}{z_1+z_2} =1 }\]
mathslover
  • mathslover
^ not necessary that : (a+bi)-(n+mi) / (a+bi)+(n+mi) = 1 ...
shubhamsrg
  • shubhamsrg
You can do it easily by interpretting is geometrically.
shubhamsrg
  • shubhamsrg
it*
amistre64
  • amistre64
im just not that attuned to complex calculations :)
AravindG
  • AravindG
|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 \(\dfrac{\pi}{2}\)
AravindG
  • AravindG
;) geometry is wonderful
mathslover
  • mathslover
wait...
mathslover
  • mathslover
can you draw that parallelogram please @AravindG
mathslover
  • mathslover
I am confused that whether that parallelogram is possible or not with two diagonals as : z1 + z2 and z1-z2
hartnn
  • hartnn
\(\huge |\frac{(a+bi)-(n+mi)}{(a+bi)+(n+mi)}|=1 \\ (a-n)^2+(b-m)^2=(a+n)^2+(b+n)^2 \\ \implies an+bm=0\) \(\huge \frac{i(a+bi)}{(n+mi)}=\frac{i(a+bi)}{(n+mi)}\dfrac{(n-mi)}{(n-mi)}= \\ \huge =\dfrac{i(an+bm-inb+ami)}{(n^2+m^2)}=\dfrac{i^2(0+am+nb)}{n^2+m^2}= \\ =k\)
mathslover
  • mathslover
well right :) I got it ..
mathslover
  • mathslover
But what about the second one..
mathslover
  • mathslover
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
AravindG
  • AravindG
@mathslover where do you have confusion in my working ?
mathslover
  • mathslover
I am confused whether that parallelogram is possible or not?
mathslover
  • mathslover
can you draw that for me please?
AravindG
  • AravindG
just consider z1 and z2 as two vectors .... P.S. I am bad at drawing
mathslover
  • mathslover
|dw:1361019499183:dw| is it right diagram ?
mathslover
  • mathslover
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)) ...
AravindG
  • AravindG
i would say this : |dw:1361019666403:dw|
AravindG
  • AravindG
gosh u made me draw that ...:P
mathslover
  • mathslover
|dw:1361019850236:dw|
mathslover
  • mathslover
is C (z_1 - z_2)
AravindG
  • AravindG
c is z1+z2
mathslover
  • mathslover
oh got it now :)
AravindG
  • AravindG
:)

Looking for something else?

Not the answer you are looking for? Search for more explanations.