Show that $\log|e^z-z|\leq c|z|$ for $|z|>R$

Question

Show that $\log|e^z-z|\leq c|z|$ for $|z|>R$.

Attempt:

$\log$ is increasing thus from the triangle inequality $$\log|e^z-z|\leq \log(|e^z|+|z|)$$ But I'm not sure how to proceed. (For very small $\Re (z)$, $|z|$ is not less then or equal to $|e^z|$).

Answer 1

Good start. Now further, $|e^z| \le e^{|z|}$, hence $$ \log|e^z-z|\leq \log(e^{|z|}+|z|) = |z|+ \log(1 + \frac{|z|}{e^{|z|}}) $$ Now $\frac{|z|}{e^{|z|}}$ has a maximum at $|z| =1$, so $$ \log|e^z-z|\leq |z|+ \log(1 + \frac{1}{e}) $$ From here, you can use the condition $|z| > R$ to compute $$ R + \log(1 + \frac{1}{e}) = R \cdot c_0(R) $$ or $$ c_0(R) = 1 + \frac{\log(1 + \frac{1}{e})}{R} $$

So for any $c \ge c_0 (R) $ and $|z| > R$, the inequality is established.$\qquad \Box$

P.S.: This might not be the tightest value for $c$.

Answer 2

Just a complementary note to @Andreas' answer. From $$\log(1+x)\leq x, \forall x> -1$$ we have $$\log\left(1+\frac{|z|}{e^{|z|}}\right)\leq \frac{|z|}{e^{|z|}}$$ leading to $$...\leq |z|+\frac{|z|}{e^{|z|}}=|z|\left(1+\frac{1}{e^{|z|}}\right)\leq |z|\left(1+\frac{1}{e^{R}}\right)$$

Answer 3

A stronger inequality is true: $|e^{z}-z|=|1+\frac {z^{2}} {2!} +\frac {z^{3}} {3!}+\cdots| \leq 1+\frac {|z|^{2}} {2!} +\frac {|z|^{3}} {3!} +\cdots \leq e^{|z|}$ for all $z$ so $\log \, |e^{z}-z| \leq |z|$ for all $z$.