Find the number of solutions of the equation $e^z = 2z+1$ in the open unit disc $\{z \in \Bbb C : |z| < 1\}$ Find the number of solutions of the equation $e^z = 2z+1$ in the open unit disc $\{z \in \Bbb C : |z| < 1\}$.
My Attempt:
Let $f(z) = e^z-2z-1$. Let $A \subseteq \{z \in \Bbb C : |z| < 1\}$ be the solution set. For solution,we will put $f(z) = 0$. Since $f(0) = 0$ so $0 \in A$. Therefore $A \neq \varnothing$. Also let $z = x+iy$ then $f(z) = 0$ gives $$e^{x+iy}-2(x+iy)-1 = 0 \implies e^xe^{iy}-(2x+1)-i(2y) =0\\
\implies e^x(\cos y + i\sin y) = (2x+1) + i(2y)$$
On comparing real and imaginary parts on both sides, we get
$$e^x\cos y = 2x + 1,\ e^x\sin y = 2y.$$
On dividing, we get $\tan y = \dfrac{2y}{2x+1}$.
I have no general formula. Please help me.
 A: If $|z|=1$, then$$\left|-e^z+1\right|\leqslant e-1<2=|2z|.$$Therefore, by Rouché's theorem, $2z$ and $2z-e^z+1$ have the same number of zeros (if we count them with their multiplicities) in $D_1(0)$. But $2z$ only has one such zero, and therefore the same thing applies to $2z-e^z+1$.
A: You already figured out that $z=0$ is one solution of the equation $e^z=2z+1$. Using the Taylor series of the exponential function and simple estimates one can show that there are no other solutions in the unit disk:
For $0 < |z| < 1$ is
$$
 e^z - (1+2z) = -z + \frac{z^2}{2!}++ \frac{z^3}{3!} + \cdots
$$
and therefore
$$
\left| \frac{e^z-(1+2z)}{z}\right| = \left|-1 + \frac{z}{2!} + \frac{z^2}{3!} + \cdots\right| \ge  \ 1- \frac{|z|}{2!} - \frac{|z|^2}{3!} - \cdots \\
\ge 1 - \frac{1}{2!} - \frac{1}{3!} - \cdots = 3 - e > 0 \, .
$$
A: Using Roche Lemma,  we mean that:
$$e^z-1-2z=0 \tag {1}  $$
We know in the circle $|z|=1$,
We have module of $2z$ is two, and module of $e^z-1$ is less than $e-1<2$,
therefore $(1)$ has as many zeros as $2z$ has in $|z|<1$.
only one zero (counting its multiplicty)
A: You have found one solution: $z=0$. Continuing from your work, for nonzero $y$, we get
$$x=\frac y{\tan y}-\tfrac12\quad\text{and}\quad x=\ln\frac{2y}{\sin y}.$$
Regarding these as functions in the form $x=f(y)$, we see that both are even (i.e. $f(y)=f(-y)$); so we need only consider solutions with $0<y\leqslant1$. In this range, the first function drops from its $x$ upper bound of $0.5$, while the second climbs from its $x$ lower bound of  $\ln2\approx0.693>0.5$. Hence there are no other solutions.
