Question

Let $z$ is a complex number such that $|z| = 3$, then the length of intercept made by locus of the point represented by $\left(-5 + \frac{9}{z}\right)$ on the x-axis is

• A. 6
• B. 3
• C. 9
• D. 12

Answer 1

Answer: (A)

6

Answer 2

Given $|z|=3$, we have $z\bar{z}=9$, which implies $\frac{9}{z} = \bar{z}$. Let $w = -5 + \frac{9}{z}$. Substituting $\frac{9}{z} = \bar{z}$, we get $w = -5 + \bar{z}$. If $w=x+iy$ and $z=a+ib$, then $x+iy = -5 + (a-ib)$, which implies $a=x+5$ and $b=-y$. Using $|z|^2=a^2+b^2=9$, we substitute $a$ and $b$: $(x+5)^2+(-y)^2=9$, so $(x+5)^2+y^2=9$. This is the equation of a circle with center $(-5,0)$ and radius $3$. To find the x-axis intercepts, set $y=0$: $(x+5)^2=9 \implies x+5=\pm 3$. The intercepts are $x=-2$ and $x=-8$. The length of the intercept is the distance between these points: $|-2 - (-8)| = 6$.