Question: Among the complex numbers $z$ satisfying $|z-25i|\le15$, the number having lowest argument is:

All the numbers so satisfy the condition.

The answer given is c) but i think the answer should be b) since it has the least argument and does satisfy the condition.

c) has an argument of $\arctan{4\over3}$ while b) has an argument of $\arctan{-5\over3}$

Am i correct or is the book correct?


You have not calculated the argument of $b$ properly. Since $b$ lies in second quadrant, its principle argument evidently lies between $\pi/2$ and $\pi$.

Principal argument of $b$ is given as the angle its radius vector makes with $+ x$ axis, and its value is $\pi+ \arctan\frac{-5}{3}$ which is greater than $\pi / 2$.

The book gives correct answer.

For more on how to calculate principle argument, refer this link.

  • $\begingroup$ So Principal argument mean the angle it makes with the positive x axis measured anticlockwise? $\endgroup$ – Anvit Jan 6 '18 at 13:18
  • $\begingroup$ Oops forgot to mention that, yes generally principal argument is taken as $(-\pi, \pi]$ so you can say for quadrants $1,2$ its measured anticlockwise and for $3,4$ its measured clockwise from $+x$ axis. $\endgroup$ – samjoe Jan 6 '18 at 13:21

A good approach to solving this question is to think about the problem geometrically.

The complex numbers $z$ satisfying the condition $|z - 25i| \leqslant 15$ is the region in the Argand plane lying on and inside a circle of radius 15 units centred at $(0,25)$.

The complex number $z$ satisfying the condition $|z - 25i| \leqslant 15$ having the least argument will geometrically be the point on the circle in the first quadrant whose tangent passes through the origin. Let us call this point $z_{\rm min}$ with principal argument $\alpha = \text{Arg} (z_{\rm min})$.

From the geometry of the problem, to find $\alpha$ we have a right-angled triangle with hypotenuse of length 25 units (the distance from the origin to the centre of the circle), side adjacent to the angle $\alpha$ of length 15 units (the radius of the circle), and opposite side of length 20 units as can be found from Pythagoras' theorem.

Thus $$\cos \alpha = \frac{15}{25} \quad \text{and} \quad \sin \alpha = \frac{20}{25} = \frac{4}{5}.$$

So $|z_{\rm min}| = 20$ (distance from the origin to the point where the tangent just touches the circle) and we have \begin{align*} z_{\rm min} &= |z_{\rm min}| \left (\cos \alpha + i \sin \alpha \right ) = 20 \left (\frac{3}{5} + \frac{4i}{5} \right ) = 12 + 16i. \end{align*} So the answer is (c).


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.