How to show this inequality for complex numbers? Let $|y|<|z|$ then, 
$$\left| \frac{y+z}{|y+z|} - \frac{z}{|z|}\right|\leq K |z|^{-1}|y|$$
for some constant $K.$
I want to use some Taylor expansion argument, but I am not sure if it would work for complex numbers. Any ideas will be much appreciated.
Edit: The expression should be equivalent to the one below:
$$\begin{align}
\left| \frac{y+z}{|y+z|} - \frac{z}{|z|}\right|&= \left| \frac{y/z+1}{|y/z+1|} - 1\right| \\
&\leq |1-y^2/z^2-1|\leq |z|^{-2}|y|^2\leq |z|^{-1}|y|
\end{align}$$
 A: Let $y=\rho e^{i\phi}$ and $z=re^{i\theta}$. Then $|y|<|z|\implies\rho<r$ and $|z|^{-1}|y|=\rho/r$. Since  \begin{align}|y+z|&=|(\rho\cos\phi+r\cos\theta)+i(\rho\sin\phi+r\sin\theta)|\\&=\sqrt{\rho^2\cos^2\phi+2\rho r\cos\theta\cos\phi+r^2\cos^2\theta+\rho^2\sin^2\phi+2\rho r\sin\theta\sin\phi+r^2\sin^2\theta}\\&=\sqrt{\rho^2+2\rho r\cos(\theta-\phi)+r^2}\end{align} we have \begin{align}\small\left| \frac{y+z}{|y+z|} - \frac{z}{|z|}\right|&=\small\left|\frac{\rho\cos\phi+r\cos\theta}{\sqrt{\rho^2+2\rho r\cos(\theta-\phi)+r^2}}+i\frac{\rho\sin\phi+r\sin\theta}{\sqrt{\rho^2+2\rho r\cos(\theta-\phi)+r^2}}-\frac{re^{i\theta}}r\right|\\&=\small\sqrt{\left(\frac{\rho\cos\phi+r\cos\theta}{\sqrt{\rho^2+2\rho r\cos(\theta-\phi)+r^2}}-\cos\theta\right)^2+\left(\frac{\rho\sin\phi+r\sin\theta}{\sqrt{\rho^2+2\rho r\cos(\theta-\phi)+r^2}}-\sin\theta\right)^2}\\&=\small\sqrt{\frac{(\rho\cos\phi+r\cos\theta)^2+(\rho\sin\phi+r\sin\theta)^2}{\rho^2+2\rho r\cos(\theta-\phi)+r^2}-\frac{2(\cos\theta(\rho\cos\phi+r\cos\theta)+\sin\theta(\rho\sin\phi+r\sin\theta))}{\sqrt{\rho^2+2\rho r\cos(\theta-\phi)+r^2}}+1}\\&=\sqrt{2-\frac{2(\rho\cos(\theta-\phi)+r)}{\sqrt{\rho^2+2\rho r\cos(\theta-\phi)+r^2}}}.\end{align} It can be shown that for $a,b,c,d\in\Bbb R^+$, the function $$f(x)=\frac{ax+b}{\sqrt{c+dx}}\implies\min f(x)=\frac{2\sqrt{a(bd-ac)}}d$$ where it exists at $x=(bd-2ac)/ad$. Substituting $a=2\rho$, $b=2r$, $c=\rho^2+r^2$, $d=2\rho r$ and $x=\cos(\theta-\phi)$, this gives \begin{align}\left| \frac{y+z}{|y+z|} - \frac{z}{|z|}\right|&\le\sqrt{2-\frac{\sqrt{2\rho(4\rho r^2-2\rho^3-2\rho r^2)}}{2\rho r}}=\sqrt{2-2\sqrt{1-\left(\frac\rho r\right)^2}}\le\sqrt2\cdot\frac\rho r.\end{align} The result follows with $K=\sqrt2$.
A: 
Found a simpler solution:

Let $y=(\lambda -1)z\iff y+z=\lambda z$ with $\lambda\neq0$ since $y\neq-z$. The inequality's left hand side then reads
$$\left|\frac{\lambda z}{|\lambda z|}-\frac{z}{|z|} \right|=\left|\frac{\lambda }{|\lambda| }-1\right|$$
whereas it's right hand side becomes
$$K|\lambda -1|.$$
We will show for any complex number $u=a+ib$ of length $1$ and any real number $t$ that
$$|u-1|^2\leq2|tu-1|^2$$
as long as the real part of $u$ is non-negative, that is $0\leq a\leq1$.  (See the geometric interpretation below.). From here the original claim follows with $u=\lambda/|\lambda|$ and $t=|\lambda|$.
We have
$$\begin{align}|u-1|^2&\leq|tu-1|^2\\
\iff
a^2-2a+1+b^2&\leq 2(t^2a^2-2at+1+t^2b^2)\\
\iff
2-2a&\leq2(t^2-2at+1)\\
\iff
0&\leq(t-a)^2+a(1-a)
\end{align}$$
Now that’s true if $0\leq a\leq1$.

Former solution:

From $|y|/|z|<1$ we know that $|\lambda -1|<1$, that is, $\lambda $ lies in the circle centred at $1$ with radius $1$, for example $\lambda =BE$, hence the length of $\lambda -1$ is the length of $AE$, makes as red line.  The inequality's left hand side is the distance between the unit vector $BD=\frac{\lambda}{|\lambda|}$ in direction of $\lambda $ and $1$,  that is the length of $AD$, marked yellow. (Notice that when you move $E$ to some other point of $BC$, the yellow line stays the same. From here we notice as well that $E$ may be beyond $C$ as well.)

We know claim that
$$\left|\frac{\lambda }{|\lambda |}-1  \right|^2\leq2|\lambda -1|^2$$
as long the real part of $\lambda$ is positive:
$$\begin{align}
\left|\frac{\lambda }{|\lambda |}-1  \right|^2&\leq2|\lambda -1|^2\\
\iff \left|\frac{\lambda }{|\lambda |}-1  \right|\cdot \left|\frac{\bar\lambda }{|\lambda |}-1  \right|&\leq2|\lambda -1|\cdot|\bar\lambda -1|\\
\iff\frac{\lambda\bar\lambda}{|\lambda|^2}-\frac{\lambda}{|\lambda|}-\frac{\bar\lambda}{|\lambda|}+1&\leq2(\lambda\bar\lambda-\lambda-\bar\lambda+1)\\
\iff 2-\frac{2\operatorname{Re}(\lambda)}{|\lambda|}&\leq2(|\lambda|^2-2\operatorname{Re}(\lambda)+1)\\
\iff0&\leq |\lambda|^2-2\operatorname{Re}(\lambda)
+\frac{\operatorname{Re}(\lambda)}{|\lambda|}
\end{align}
$$
With $\lambda=r\cdot e^{i\phi}$ the right hand side becomes
$$\begin{align}r^2-2r\cos(\phi)+\cos(\phi)&=\bigl(r-\cos(\phi)\bigr)^2+\cos(\phi)-\cos^2(\phi)\\
&=\bigl(r-\cos(\phi)\bigr)^2+\cos(\phi)\bigl(1-\cos(\phi)\bigr),
\end{align}$$ 
which certainly holds if $-\pi/2<\phi<\pi/2$, that is, the real part of $\lambda$ is positive.
