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Question: Among the complex numbers $z$ satisfying $|z-25i|\le15$, the number having lowest argument is:
a)$10i$
b)$-15+25i$
c)$12+16i$
d)$7+12i$

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?

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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.

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  • $\begingroup$ 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
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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).

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