Question

Among the complex numbers which satisfies $|z-3+3i|=3$, then the complex number having least positive argument is

• A. 3
• B. 3i
• C. -3i
• D. None of these

Answer 1

Answer: (C)

-3i

Answer 2

We start with the given equation

$$|z-3+3i|=3.$$

Writing

$$z=x+yi,$$

we note that

$$z-3+3i = (x-3)+i(y+3).$$

Thus the locus is

$$(x-3)^2+(y+3)^2=9,$$

i.e. a circle with center $(3,-3)$ and radius $3$.

A point $z$ on the circle has argument $\theta=\arg(z)$. In the standard principal value (i.e. $-\pi < \theta \le \pi$), note that the circle’s highest (largest $y$) point is $(3,0)$ having argument $0$. But since $0$ is not positive, the “least positive argument” must be taken from the representation where arguments are given in $[0,2\pi)$. In that interval the point $(3,0)$ has argument $0$ (not positive) and every other point on the circle will have an argument (measured counterclockwise) larger than $0$.

Now, geometrically, the circle lies in the fourth quadrant (with $y\le 0$) except the topmost point. In the $[0,2\pi)$ description a point in the fourth quadrant gets an argument

$$\theta = 2\pi -\alpha, \quad \alpha>0.$$

To have the least positive angle (i.e. the smallest number in $(0,2\pi)$), we need $\alpha$ as large as possible. Inspection shows that when $x=0$ the point is

$$(0,y) \quad\text{with}\quad (0-3)^2+(y+3)^2 = 9.$$

Solving:

$$9+(y+3)^2=9\implies (y+3)^2=0\implies y=-3.$$

That is, $z=-3i$. In $[0,2\pi)$ notation its argument is

$$\theta=270^\circ= \frac{3\pi}{2},$$

and checking other points on the circle gives arguments larger than $270^\circ$. (Note that the point $z=3$ with argument $0$ is excluded from “positive”.)