Question

If z=x+iy and imaginary part of (2z+1)/(iz+1) is -2,then find the value of x+2y-2.​

Answer 1

✪ᴀɴsᴡᴇʀ✪

• $\boxed{x + 2y - 2 = 0}$

★ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ★

Here

$\:\:\:\:\:\:\:\:\:\:\: Im\left[\frac{2z+1}{iz + 1} \right] = - 2$

$\implies Im\left[\frac{2(x+iy)+1}{i(x+iy)+ 1}\right] = - 2$

$\implies Im\left[\frac{2x+1+2yi}{ix - y + 1}\right] = - 2$

$\implies Im\left[\frac{2x+1+2yi }{1-y+ ix}\right] = - 2$

$\implies Im\left[\frac{(2x+1+2yi)(1-y- ix) }{(1-y+ ix)(1-y - ix)}\right] = - 2$

$\implies Im\left[\frac{(2x+1)(1-y) - (2x+1)xi +2y(1-y)i + 2xy)}{(1-y)^2 + x^2}\right]$

$\:\:\:\:\:\:\:\:\:\:= - 2$

$\implies Im\left[\frac{(2x+1)(1-y) + 2xy}{(1-y)^2 + x^2} + i\frac{2y(1-y) - (2x+1)x}{(1-y)^2 + x^2}\right]$

$\:\:\:\:\:\:\:\:\:\:= - 2$

$\implies \left[\frac{2y(1-y) - (2x+1)x}{(1-y)^2 + x^2}\right] = - 2$

$\implies 2y - 2y^2- 2x^2 - x$

$\:\:\:\:\:\:\:\:\:\:=-2[ 1 - 2y + y^2 + x^2]$

$\implies 2y - 2y^2- 2x^2 - x$

$\:\:\:\:\:\:\:\:\:\:= -2 + 4y - 2y^2 - 2x^2$

$\implies 2y - x = -2 + 4y$

$\implies 0 = -2 + 4y - 2y + x$

$\implies x + 2y - 2 = 0$

Thus, it represents a straight line in the complex plane.