Question

If a complex number $z = x+iy$ represents a point $P(x, y)$ in the Argand plane and z satisfies the condition that the imaginary part of $\frac{z-3}{z+3i}$ is zero, then the locus of the point P is

• A. $x^2+y^2-3x+3y= 0, (x, y) \neq (0,-3)$
• B. $2xy-3x+3y+9=0, (x, y) \neq (0,-3)$
• C. $x-y-3=0, (x, y) \neq (0,-3)$
• D. $x+y+3=0, (x, y) \neq (0,-3)$

Hint:

A number is purely real if its imaginary part is zero. For a complex fraction $\frac{z_1}{z_2}$, a quick way to find the condition for it to be purely real is to set $\text{Im}(z_1\bar{z_2}) = 0$. In this case, $z_1 = z-3$ and $z_2 = z+3i$.

Answer 1

The Correct Option is C

Let the complex number be $z = x+iy$.
We are given the expression $\frac{z-3}{z+3i}$. Let's substitute $z=x+iy$.
$\frac{(x+iy)-3}{(x+iy)+3i} = \frac{(x-3)+iy}{x+i(y+3)}$.
To find the imaginary part, we first rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $x-i(y+3)$.
$\frac{(x-3)+iy}{x+i(y+3)} \times \frac{x-i(y+3)}{x-i(y+3)} = \frac{[(x-3)x + y(y+3)] + i[yx - (x-3)(y+3)]}{x^2 + (y+3)^2}$.
The imaginary part of this expression is $\frac{yx - (x-3)(y+3)}{x^2 + (y+3)^2}$.
We are given that the imaginary part is zero.
So, $yx - (x-3)(y+3) = 0$.
$yx - (xy + 3x - 3y - 9) = 0$.
$yx - xy - 3x + 3y + 9 = 0$.
$-3x + 3y + 9 = 0$.
Dividing the equation by $-3$, we get:
$x - y - 3 = 0$.
The locus is a straight line. Also, the denominator of the original expression cannot be zero, so $z+3i \neq 0$, which means $x+iy \neq -3i$. Thus, $(x,y) \neq (0,-3)$.