Discussion Forum



1)

Let a,b,x and y be real numbers such that a-b=1 and y≠ o. If the complex number z=x+iy satisfies  $Im(\frac{az+b}{z+1})=y$ , then which of the following is(are) possible value(s) of x?


A) $1-\sqrt{1+y^{2}}$

B) $-1-\sqrt{1-y^{2}}$

C) $1+\sqrt{1+y^{2}}$

D) $-1+\sqrt{1-y^{2}}$

Answer:

Option B,D

Explanation:

$\frac{az+b}{z+1}=\frac{ax+b+aiy}{(x+1)+iy}$

$= \frac{(ax+b+aiy)((x+1)-iy)}{(x+1)^{2}+y^{2}}$

$\because$   $Im(\frac{az+b)}{z+1})=\frac{-(ax+b)y+ay(x+1)}{(x+1)^{2}+y^{2}}$

$\Rightarrow$     $\frac{(a-b)y}{(x+1)^{2}+y^{2}}=y$

$\because$    a-b=1

$\because$   $(x+1)^{2}+y^{2}=1$

$\because$    $x=-1\pm \sqrt{1-y^{2}}$