If-the-imaginary-part-of-2z-1-iz-1-is-2-then-the-locus-of-the-point-representing-z-in-the-complex-plane-is-

Question Number 19734 by Tinkutara last updated on 15/Aug/17

If the imaginary part of \frac{2 z + 1}{i z + 1} is - 2,

then the locus of the point representing

z in the complex plane is

Answered by ajfour last updated on 15/Aug/17

\frac{2 z + 1}{iz + 1} = \frac{(2 z + 1) (1 - i \bar{z})}{(iz + 1) (1 - i \bar{z})}

= \frac{2 z - 2 i ∣ z^{2} ∣ + 1 - i \bar{z}}{iz + ∣ z ∣^{2} + 1 - i \bar{z}}

if z = x + iy

= \frac{2 (x + iy) - 2 i (x^{2} + y^{2}) + 1 - ix - y}{1 + x^{2} + y^{2} - 2 y}

as imaginary part of above complex

number is = - 2, we can write

\frac{2 y - {2 x}^{2} - {2 y}^{2} - x}{1 + x^{2} + y^{2} - 2 y} = - 2

or 2 y + x - 2 = 0

\Rightarrow 4 iy + 2 ix - 4 i = 0

or 2 (z - \bar{z}) + i (z + \bar{z}) - 4 i = 0

(2 + i) z - (2 - i) \bar{z} - 4 i = 0

\Rightarrow (1 - 2 i) z + (1 + 2 i) \bar{z} + 4 = 0 .

Commented by Tinkutara last updated on 15/Aug/17

Thank you very much Sir!