Question

Let w be a complex number.Then the set of all complex numbers z satisfying the equation
w-w1z =k(1-z) for some real number k
Here w1 is conjugate of w...

Answer 1

@parthkohli

Answer 2

@priyar

Answer 3

@faiqraees

Answer 4

@phi

Answer 5

how far did you get ?

Answer 6

I assumed that w is a+ib and z is x+iy...
Then i solve the LHS n RHS...as per the equation

Answer 7

I got this..
a-ax-by=k(1-x)
b+bx-ay=-ky

Answer 8

I think you get
\[ z = \frac{k-w}{k-w'}\]
now I'm trying to figure out what that means (perhaps geometrically?)

Answer 9

or z= (w-k)/(w'-k)
(a-k) + bi / (a-k) - bi

in polar coords, they both have the same magnitude
so z will have magnitude 1

Answer 10

I m getting it as z=(-1)e^itheta
Theta is argument for w

Answer 11

I'm would think it's 2 * theta
z= exp(i theta) / exp(-i theta) = exp(i 2 theta)

Answer 12

wait, my theta is the arg of < (a-k) , b>

Answer 13

Yep....U r very correct....@phi ur method of modulus was apt...

Answer 14

Same moduli of the numerator n denominator gives us 1

Answer 15

Which would serve the modulus of z

Answer 16

Thanks...@phi

Answer 17

I think as you vary k, we can generate any angle we want, so the solutions will lie on the unit circle. But double check my thinking.