In complex analysis: Let $S = \{ z : z = 0 \text{ or } |\text{Arg } z| \leq \alpha \}$, where $0 < \alpha < \pi$. Verify that the function $f(z) = \sqrt{z}$ is uniformly continuous on $S$, starting by showing that $|f(z) - f(w)| \leq c|z - w|$ for $z$ and $w$ in $S$ satisfying $|z| \geq 1$ and $|w| \geq 1$, where $c = 2^{-1}\sec(\alpha/2)$. Then, focus on the compact set $S \cap \overline{B}(0, 2)$.\
I have tried to solve it in several ways but it is quite complicated or I can't find how to get to the constant c.
Note that $$\sqrt{z}-\sqrt{w} = \frac{z-w}{\sqrt{z}+\sqrt{w}}.$$
Suppose $z = r_1 e^{i\theta_1}$ and $w=r_2e^{i\theta_2}$. [$\theta_1=Arg(z)$ and $\theta_2 =Arg(w)$.] Notice that $\sqrt{z}$ and $\sqrt{w}$ lie either in first or fourth quadrants and hence their real parts are positive. Thus $$|\sqrt{z}+\sqrt{w}| \geq \Re(\sqrt{z}) + \Re(\sqrt{w}).$$ That is
$$|\sqrt{z}+\sqrt{w}| \geq r_1^{1/2}\cos\left(\frac{\theta_1}{2}\right)+r_2^{1/2}\cos\left(\frac{\theta_2}{2}\right).$$
As both $r_1$ and $r_2$ are greater than or equal to $1$ and as both $\cos\left(\frac{\theta_1}{2}\right)$ and $\cos\left(\frac{\theta_2}{2}\right)$ are positive, we must have
$$|\sqrt{z}+\sqrt{w}| \geq \cos\left(\frac{\theta_1}{2}\right) + \cos\left(\frac{\theta_2}{2}\right)$$
But
$\Big|\frac{\theta_1}{2}\Big| \leq \frac{\alpha}{2}$ & $\Big| \frac{\theta_2}{2}\Big| \leq \frac{\alpha}{2}$ and also $0 <\frac{\alpha}{2}< \frac{\pi}{2}$. Thus we must have $$\cos\left(\frac{\theta_1}{2}\right) \geq \cos\left(\frac{\alpha}{2}\right)$$ and $$\cos\left(\frac{\theta_2}{2}\right) \geq \cos\left(\frac{\alpha}{2}\right).$$ Thus $$|\sqrt{z}+\sqrt{w}| \geq 2 \cos\left(\frac{\alpha}{2}\right).$$
This shows that
$$|\sqrt{z}-\sqrt{w}| \leq \frac{|z-w|}{2\cos\left(\frac{\alpha}{2}\right)} = 2^{-1}\sec\left(\frac{\alpha}{2}\right)|z-w|.$$
Hope, you can continue with this.