Prove that the exponential $\exp z$ is not zero for any $z \in \Bbb C$

Question

How can the following been proved?

$$ \exp(z)\neq0, z\in\mathbb{C} $$

I tried it a few times, but I failed. Probably it is extremly simple. If a draw the unit circle and then a complex number $\exp(a+ib)=\exp(a)\exp(ib)$ then it is obvious that this expression is only $0$ if $\exp(a)$ equals zero, but $\exp(a),a\in\mathbb{R}$ is never zero. This seems not to be very robust.

Thank you

Answer 1

If you know that $\exp(z+w)=\exp(z) \exp(w)$, then $\exp(z)\ne 0$ follows from $$1=\exp(0)=\exp(z-z)=\exp(z)\exp(-z)$$

Answer 2

Using the given definition, $$\exp z := \sum_{k = 0}^{\infty} \frac{z^k}{k!},$$ we have $$\exp z \exp (-z) = \sum_{k = 0}^{\infty} \frac{z^k}{k!} \sum_{l = 0}^{\infty} \frac{(-z)^l}{l!} = \sum_{k = 0}^{\infty} \sum_{l = 0}^{\infty} \frac{(-1)^l z^{ k + l}}{k! l!} .$$ Absolute convergence permits us to reorder the double summation in the second equality above, and also allows us to write it via reindexing as $$\sum_{p = 0}^{\infty} \sum_{q = 0}^p \frac{(-1)^q z^p}{(p - q)! q!} .$$ Substituting gives that the $p = 0$ term is $1$. For $p > 0$ the coefficient of $z^p$ (in the outer sum) is, by definition of $\cdot \choose \cdot$, $$\sum_{q = 0}^p \frac{(-1)^q}{(p - q)! q!} = \frac{1}{p!}\sum_{q = 0}^p {p \choose q} 1^{p - q} (-1)^q .$$ We've written the sum on the r.h.s. so that we can recognize it as the binomial expansion for $(1 - 1)^p = 0$, so for $p > 0$ the coefficient of $z^p$ in the series is $0$, hence $$\exp z \exp(-z) = 1 .$$ In particular, there is no $z$ such that $\exp z = 0$.

Answer 3

$\exp(z)\neq 0 $ $\space , \forall z\in \Bbb{C}$

Proof:

$f(z) =\exp(z) $ is an entire function i.e holomorphic on $\Bbb{C}$.

To show : $\scr{Z_f}=\{z:f(z)=0\}=\emptyset$

We argue by contradiction. Suppose$\scr{Z_f}\neq\emptyset$.

Let, $z_0\in \scr{Z_f}$

Then, $f^n{(z_0)}=\exp(z_0) =0$ $ \space, \forall n\in \Bbb{N}$

Since, $f\in H(\Bbb{C})$, then $f$ has a Taylor Series about the point $z_0$ converges for all $z\in \Bbb{C}$.

$f(z) =\sum_{n=0}^{\infty}\frac{f^n(z_0) }{{n!}}(z-z_0)^n$$\space , \forall z\in \Bbb{C}$

Hence, $f(z) =0$ on $\Bbb{C}$.

But, $f(0) =\exp(0) =1$ ,$\space $ a contradiction.

Hence, $\scr{Z_f}=\{z:f(z)=0\}=\emptyset$

In other words, $\exp(z) \neq 0$ for any $z\in \Bbb{C}$

Answer 4

Let’s prove that ez=0 has no solution ∀z∈C

We are going to use the fact that if f:C⟶C is holomorphic on the whole C plane, then it is equal to its taylor series on the whole C.

First, let’s prove that f(z)=ez is holomorphic on the whole C using the Cauchy-Riemann Equations, which state that a complex function f(z)=u(x,y)+v(x,y)i is holomorphic if and only if the following conditions are satisfied:

∂u∂x=∂v∂y

∂u∂y=−∂v∂x

To write our function f(z)=ez separating the real and imaginary parts, we use Euler’s formula:

f(z)=ez=ex+iy=ex(cos(y)+isin(y))=excos(y)+iexsin(y)

Using Cauchy-Riemann Equations:

∂u∂x=excos(y)=∂v∂y

and

∂u∂y=−exsin(y)=−∂v∂x

Since Cauchy-Riemann Equations are satisfied and the partial derivatives are continuous, we can conclude that the function is holomorphic ∀z∈C

Now, how could we proceed? Well, as I said if a function is holomorphic on the whole complex plane, than all the functional values on the plane are equal to the function’s taylor series. Now, let’s assume:

∃z0∋ez0=0

Then, since all the derivatives of ez are itself, all the derivatives are zero at z0 , which implies that the Taylor series of the function centered at z0 is identically 0 in the whole complex plane, which is clearly a contradiction. Hence, there doesn’t exist any z∈C∋ez=0