Some inequality with complex numbers Is there a straightforward way to prove the following inequality:
$$|1 + k\big(\exp(it)-1\big)|\leq 1 $$
where $k\in(0,1)$ and $t \in \mathbb{R}$ (correction, see dxiv answer) with $|t| \leq 1$, other than writing the quantity into its real and imaginary parts and checking that they satisfy the required inequalities (which is long and seems inelegant) ?   
 A: The inequality does not hold true for $t \in \mathbb{C}$ with $|t| \leq 1\,$ in general, take for example $t=-i$ then $\,e^{it}=e\,$ and $\,|1 + k (e-1)|=1+k(e-1) \gt 1\,$ since $\,e \gt 1$ and $k \gt 0\,$.
Assuming $t \in \mathbb{R}\,$, instead, let $z=e^{it}\,$, then $|z|=1\,$ and:
$$
\begin{align}
|1 + k (z-1)|^2 &= \big(1 + k (z-1)\big)\big(1 + k (\bar z-1)\big) \\
 &= 1 + k(\bar z -1) + k(z-1) +k^2(z-1)(\bar z - 1) \\
 &= 1 + k(z+\bar z) -2k +k^2(z \bar z-z - \bar z + 1) \\
 &= 1 + (k-k^2)(z+\bar z) - 2k+2k^2 \\
 &= 1 + k(1-k)\cdot 2 \operatorname{Re}(z) - 2k(1-k) \\
 &= 1 - 2k(1-k)\big(1-\operatorname{Re}(z)\big) \\
 &\le 1 \quad\quad \text{since} \;\;k(1-k) \ge 0 \;\;\text{and} \;\;\operatorname{Re}(z) \le |z| = 1
\end{align}
$$
A: Look at it geometrically. Fixing $k$, the left side ranges over the circle with center $1-k$ and radius $k$. We see that this circle is internally tangent to the unit circle, so regardless of $k$ any point on the LHS must be contained in the unit circle. This means they are of magnitude $\leq 1$. 
