How can i prove $h(x)>g(x)$ when $0I'm trying to prove that $h(x) > g(x)$ when $0<x<1$.
$$h(x) = (1 + i \cdot x)$$
$$g(x) = (1 + i)^x$$
And $i$ is a real number greater than $0$.
I tried with two differents ways, but i wasn't successful in none of them.. 
First attempt:
$$(1+ i \cdot x) \;\textrm{   ... }\; (1+i)^x$$
$$ \ln{(1 + i \cdot x)} \;\textrm{   ... }\; x \cdot \ln{(1+i)}$$
$$ 1 \;\textrm{   ... }\; \dfrac{x \cdot \ln{(1+i)}}{\ln{(1 + i \cdot x)}}$$
So i have to observe the function $f(x) = \dfrac{x \cdot \ln{(1+i)}}{\ln{(1 + i \cdot x)}}$ in interval $0<x<1$ to determine if it's greater, equal or smaller than 1.
First i have to know if $f(x)$ has a critical point in this interval. 
$$f '(x) =  \dfrac{\ln{(1+i)}}{\ln{(1+i \cdot x)}} - \dfrac{i \cdot\ln{(1+i) \cdot x}}{(i\cdot x +1)\cdot \ln{(i\cdot x +1)^2}}$$
I tried to find the roots (if exit roots, but couldn't resolve..)
So i left this attempt, and tried in a different way:
Second Attempt:
I know that $h(x)$ is a linear function and $g(x)$ is a convex function. If i prove $g(x)$ is convex, and $h(x)$ is a linear combination of $g(x)$ in interval $0<x<1$, then $h(x) > g(x)$ in this interval, alright?
From definition to a function be convex:
$$ g(t \cdot a + (1-t)\cdot b)\leq t\cdot g(a) + (1-t) \cdot g(b) \textrm{ , where } t \in [0,1]$$
$$ (1+i)^{(t\cdot a + (1-t)\cdot b)} \leq t\cdot (1+i)^a + (1-t)\cdot (1+i)^b$$
I'm stucked here..
Any idea how can i prove $h(x)>g(x)$??
 A: I think you can define $f(x) = h(x) - g(x) = 1+ix - (1+i)^x$. Then $f^{'} (x) = i - (1+i)^x \ln (1+i)$.
We have $f^{'}(x) = 0$ iff $x = \dfrac{\ln i - \ln \ln (1+i)}{\ln (1+i)}$. Note that $e^i > 1 + i$, hence $x_0 = \dfrac{\ln i - \ln \ln (1+i)}{\ln (1+i)} > 0$.
On the other hand, it is easy seen that $f^{'}(x) < 0$ when $1 > x > x_0$ and $f^{'}(x) > 0$ when $0 < x < x_0$. Since $f(0) = f(1) = 0$, we obtain $f(x) \geq 0$ when $x \in [0,1]$. This mean that $h(x) \geq g(x)$ if $x \in [0,1]$.
A: By observation we see that $h(0) = g(0)$ and $h(0) = g(0)$
Let $f(x)=\frac{h(x)}{g(x)}$ then if the postulate $h(x) > g(x)$ is true for $0<x<1$ then $f(x)>1$ for this domain of $x$. As $g(x)>0$ for $x>0$, $f(x)$ is always defined.
Initially we have $f(0) = 1$ and $f(1) = 1$
Then differentiating with respect to x; $$f'(x) = \frac{p(1+p)^x - (1+px)(\ln(1+p))(1+p)^x}{(1+p)^{2x}}$$
$$f'(x) = \frac{(1+p)^x[p - \ln(1+p) - p \ln(p+1)x]}{(1+p)^{2x}}$$
As $f'(x) = 0$ at stationary points;
$$ p - \ln(p+1) - p \ln(p+1)x = 0 $$
therefore $$x=\frac{p - \ln(p+1)}{p \ln(p+1)}$$
As $p > \ln(p+1)$ for all $p \in$ $\Re$ therefore  $$0 < \frac{p - \ln(p+1)}{p \ln(p+1)} <1$$ checking the gradient of the graph we see that a maxima occurs a the above value.
As $f(0) = 1$ and $f(1) = 1$ and the graph is increasing upto $\frac{p - \ln(p+1)}{p \ln(p+1)}$ we can say for certain that $f(x) > 1$ for $0<x<1$ 
Thereby for $0<x<1$ , $$\frac{h(x)}{g(x)} > 1$$
Therefore for $0<x<1$ $$ h(x) > g(x)$$
