Prove that $\sum_{n=0}^{\infty}(-1)^{n}z^{2n}/({2n})! $ is convergent for all $z ∈ \mathbb{C}$
I am familiar how to tackle this if it were a Real Analysis problem, but unsure how using Leibniz's test changes the result given that it's Complex.
I see how this is $cos(z)$ but how I don't see how to explicitly prove that it is convergent using conventional methods.
The ratio and root test apply as normal. Indeed, from these tests we define the radius of convergence of a power series $\sum a_nx^n$ as
$$\rho:=\frac1{\limsup_{n\to\infty}\sqrt[n]{|a_n|}}$$
and when the limit exists
$$\rho=\lim_{n\to\infty}\frac{|a_n|}{|a_{n+1}|}$$
This definition is directly derived from these tests when you clear the $x$. The radius mean that the series converges if $|x|<\rho$. Sometimes it converges too when $|x|=\rho$. Take a look at this.
Well let's have ($a_n$) a complex sequence verifying: $|\frac{a_{n+1}}{a_n}| \rightarrow 0$ , with $a_n \neq0 $ ,for n integer
Then for a sufficient N, we have: $n>N \implies |\frac{a_{n+1}}{a_n}|<\frac{1}{2}$
Hence you can prove that : $|\frac{a_{n}}{a_N}| < \frac{1}{2^{n-N}} $ , for any n>N
Which means ($\sum a_n$) converges. Of course this stays true for a series of functions, given that the ratio condition is true for every z.
So you can use that