Find one $z\in \mathbb{C}$ in the inequality $|z-25i|\le 15$ that has the largest argument ($\arg (z)$) Find one $z\in \mathbb{C}$ in the inequality $|z-25i|\le 15$ that has the largest argument ($\arg (z)$)
The inequality is equivalent to $x^2+(y-25)^2\le 15^2$ that represents the set of points in the circle of radius $15$ and center coordinate $C(0,25)$.
In this set, how to find one complex number which has the largest argument?
 A: Geometric Approach
The image below shows how to compute the point in the given region with the greatest argument.


Calculus Approach
The argument is an increasing function of $\frac yx$ when $y\gt0$. Therefore, we need to maximize $\frac yx$ on the boundary, which can be parametrized as
$$
(x,y)=(15\cos(\theta),25+15\sin(\theta))\tag{1}
$$
That is, maximize
$$
\frac{25+15\sin(\theta)}{15\cos(\theta)}=\frac53\sec(\theta)+\tan(\theta)\tag{2}
$$
whose derivative is
$$
\frac53\tan(\theta)\sec(\theta)+\sec^2(\theta)\tag{3}
$$
multiply by $\cos^2(\theta)$ to get
$$
\frac53\sin(\theta)+1\tag{4}
$$
which vanishes when $\sin(\theta)=-\frac35$, and since $x\lt0$, $\cos(\theta)=-\frac45$. Plug these values back into $(1)$ to get
$$
(x,y)=(-12,16)\tag{5}
$$
A: Having posed that $z = x+iy$, then:
$$|z-25i|\le 15 \Rightarrow x^2 +(y-25)^2 \le 15^2.$$
Recall that:
$$\arg(z) = 
\begin{cases}
f(y) & x > 0 \\
f(y) + \pi & x<0 \wedge y \ge 0 \\
f(y) - \pi & x<0 \wedge y < 0\\
\frac{\pi}{2} & x = 0 \wedge y > 0 \\
-\frac{\pi}{2} & x = 0 \wedge y < 0 \\
\text{undefined} & x = 0 \wedge y = 0
\end{cases},$$
where 
$$f(y) = \arctan\left(\frac{y}{\sqrt{15^2 - (y-25)^2}}\right).$$
Notice that $x = \pm \sqrt{15^2 - (y-25)^2}$.
We want to find the maximum of $\arg(z)$ with respect to $y$. The $\arctan$ is monotonically increasing, then we can work on $f(y)$:
$$\frac{\partial }{\partial y} f(y) = \frac{25(y-16)}{\left(\sqrt{15^2 - (y-25)^2}\right)^3}.$$
Imposing it equal to $0$, you get $y = 16$ and $x=12$.
For $z = 12 + i16$, the argument is:
$$\arg(z) = \arctan\left(\frac{16}{\sqrt{15^2 - (16-25)^2}}\right) \simeq 0.9273.$$
When $x$ is negative, that is $x = - \sqrt{15^2 - (y-25)^2}$, then we have another candidate $z = -12 + 16i$. In this case $\arg(z) = \arctan\left(\frac{16}{\sqrt{15^2 - (16-25)^2}}\right) + \pi \simeq 4.0689$.
Last case to consider is $x = 0$ and $y = 40$. Here, we get that $\arg(z) = \frac{\pi}{2} \simeq 1.5708.$ 
Finally, we conclude that $z = -12+16i$ is the maximum of $\arg(z)$, and $z$ stays in the circle you described.
A: It is clear the largest argument corresponds to the red tangent below. The involved triangle is $5$ times the pythagorean one of sides $(3,4,5)$ i.e. of sides $(15,20,25)$. Thus $$z=-12+16i$$

