Among the complex numbers $z$ which satisfies $|z-25i|\leq 15$, find the complex number $z$ having:

(A) Least positive argument

(B) Greatest positive argument

(C) Least modulus

(D) Greatest modulus

As $|z-25i|\leq 15$ represents the boundary and interior of a disk, I think the maximum and minimum modulus can be found along the line joining centre and origin. But I am not able to think about argument part. What should be the approach in that case?


Hint The argument is constant on a ray through the origin, and increases as the ray sweeps anticlockwise (until it intersects the branch cut, anyway). Thus (assuming the branch cut is tucked suitably away from the circle), if the argument is minimal/maximal at a point on the circle, the constant-argument ray is tangent to the circle there. (Otherwise, we could rotate the ray a little further from the center of the circle and decrease/increase the argument.)

Now, any radius of a circle is orthogonal to the tangent line to the circle at the point where it intersects that radius.


To answer the part about the arguments, we can construct tangent lines from origin to circle.

There should be two of them.

Let consider the tangent that intersects the circle in the first quadrant. By definition of tangent, if you rotate this line a little bit to the right, you are not going to intersect the disk. If you rotate this line a little bit to the left, you are going to miss some portion of the disk. Since argument increases in anti-clockwise direction, the angle form by this tangent line and the x-axis is the minimal argument.

Let the intersection point be $Q$. Let $P=(0,25)$ and let $O$ be the origin.

$\Delta PQO$ is a right triangle. We can easily use trigonometry to compute $\angle POQ$. $$\angle POQ= \arcsin\frac{15}{25}$$

and minimal argument $=\frac{\pi}{2}-\angle POQ.$

By symmetry, we can see that maximal argument = $\pi$-minmal argument.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.