How was this function approximated? I found this passage in my book which is not very clear to me as to how it was obtained. 
Given $n\in\mathbb{N},\,k\in(0,1)$, then
$$(1+i)^{n+k}\approx(1-k)(1+i)^n+k(1+i)^{n+1}=(1+i)^n(1+ki)$$
 A: Is $i$ supposed to be small? If so, the derivative of $(1+i)^{n}$ with respect to $n$ is $(1+i)^n\log(1+i)$. Thus a first order approximation of $(1+i)^{n+k}$ is 
$$
(1+i)^{n}+k(1+i)^{n}\ln(1+i). 
$$
If $0<i<1$ then $i\approx\ln(1+i)$ (again using a Taylor series approximation), and hence one arrives at the desired approximation of $(1+i)^{n}+ik(1+i)^{n}$.
A: Some intuition to go with ThomasGrubb's thorough answer: for $a>1$, $a^x$ is an increasing function that interpolates between $1$ (at $x=0$) and $a$ (at $x=1$). A secant approximation of $a^x$ is ordinary linear interpolation of these extremes: $(1-x)\cdot 1 + x\cdot a.$
As ThomasGrubb notes, this approximation is increasingly poor as $a$ increases.
A: Notice that $k$ is assumed between $0$ and $1$. With $f(k):=(1+i)^{n+k}$, the approximation is that $f(k)$ grows roughly linearly as $k$ ranges from $0$ to $1$. Consequently,
$$
\frac{f(k)-f(0)}{k-0}\approx\frac{f(1)-f(0)}{1-0}
$$
Rearranging this gives
$$
f(k)\approx (1-k)f(0) + k f(1).
$$
