Consider the complex function $f(z)=\cos z+e^{z^2}$. The coefficient of $z^5$ in the Taylor series expansion of $f(z)$ about the origin is __________ (rounded off to 1 decimal place).

The correct answer is 0.

To find the coefficient of $z^5$ in the Taylor series expansion of the complex function $f(z)=\cos z+e^{z^2}$ about the origin, we can analyze the Taylor series expansion of each term individually.

First, recall the Taylor series expansion for $\cos z$ about the origin (which is its Maclaurin series):

$\cos z=1-\frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}+...=\sum _{n=0}^{\infty }\frac{(-1{)}^n{z}^{2n}}{(2n)!}$

Note that the expansion of $\cos z$ contains only even powers of $z$. Therefore, the coefficient of any odd power of $z$ (such as $z^5$) in this expansion is 0.

Next, let's look at the Taylor series expansion for $e^{z^2}$ about the origin. We start with the standard exponential series:

$e^w=1+w+\frac{w^2}{2!}+\frac{w^3}{3!}+...=\sum _{n=0}^{\infty }\frac{w^n}{n!}$

Substituting $w=z^2$ into this series gives:

$e^{z^2}=1+z^2+\frac{z^4}{2!}+\frac{z^6}{3!}+...=\sum _{n=0}^{\infty }\frac{z^{2n}}{n!}$

Just like $\cos z$, the expansion of $e^{z^2}$ contains only even powers of $z$ (powers of the form $2n$). Consequently, the coefficient of any odd power of $z$ (including $z^5$) in this expansion is also 0.

Since $f(z)$ is the sum of these two functions, its Taylor series expansion is the term-by-term sum of their individual series:

$f(z)=(1-\frac{z^2}{2!}+\frac{z^4}{4!}-...)+(1+z^2+\frac{z^4}{2!}+...)$

Since neither series contains a term with $z^5$, the coefficient of $z^5$ in the overall expansion of $f(z)$ is:
$0+0=0$