Smallest $k$ such that if $z= x + iy$ , then $|x| + |y| \leq k|z|$ If $z= x + iy$, then $|x| + |y| \leq k|z|$, where the smallest possible value of $k$ is?
Options:
$a) \ 1$
$b)\  \sqrt 2$
$c)\ \sqrt3$
$d)$ None of the above.

By using geometry, I can see that if $|x|$ is equal to $|y|$, then $|z|$ will become equal to $\sqrt2|x|$ (or $\sqrt 2|y|$). But, I would like a mathematical proof for the same which I am unable to arrive at. Any help is welcomed.
 A: Rewriting $z$ in polar form, let $z = |z| \cos \theta + i|z|\sin\theta$.
If $z$ is in the first quadrant, $0 \le \theta \le \frac\pi2$, then
$$\begin{align*}
|x| + |y| &= x + y\\
&= |z| \cos\theta + |z|\sin\theta\\
&= \sqrt2 |z| \cos\left(\theta - \frac\pi4\right)\\
&\le \sqrt 2|z|
\end{align*}$$
Equality holds when $\theta = \frac\pi4$, i.e. $x = y$.
The result is similar if $z$ is in other quadrants, but where $|x|$ may equal to $-x$ instead, or $|y|$ may equal to $-y$ instead.
A: As you already noted if  $|x|=|y|=1$ then $|z|=\sqrt{2}$ and 
$$2=|x| + |y| \leq k|z|=k\sqrt{2}\implies k\geq \sqrt{2}.$$
In order to show that the smallest constant $k$ such that the  inequality $|x| + |y| \leq k|z|$ holds is $\sqrt{2}$, it suffices to prove that
$$(|x|+|y|)^2\leq 2|z|^2=2(x^2+y^2),$$
which is true because 
$$2(x^2+y^2)-(|x|+|y|)^2=|x|^2-2|x||y|+|y|^2=(|x|-|y|)^2\geq 0.$$
A: Assuming $x $ and $y$ are real.
Let $Z=x +iy =re^{i\theta}=r(\cos{\theta}+i\sin{\theta})$
$$|x|+|y|=r(|\cos{\theta}|+\sin{\theta}|)$$
$$(|x|+|y|)^2=r^2(1+|\sin{2\theta}|)$$
Now use $|\sin{x}|\le 1$
