How do I express $e^{polynomial}$ in complex numbers

Question

Consider the polynomial $Q_{\alpha}(x)=x^3+\alpha x+1$.

Express the complex number $e^{i Q_{-1}(i)}$ in the form $a+bi$ where $a,b \in \mathbb{R}$.

The answer is $e^2 \cos 1 +i e^2 \sin1$.

How do I get there?

Answer 1

You have your polynomial: $$Q_{\alpha}(x)=x^3+\alpha x+1$$ Therefore, if we substitute the value $\alpha=-1$ and $x=i$, we obtain:

$$Q_{-1}(i)=i^3-i+1$$

Note that $i^3=-i$

Hence,

$$Q_{-1}(i)=-2i+1 \tag{1}$$

Therefore you can substitute $(1)$ in your main equation:

$$e^{i Q_{-1}(i)}=e^{i \cdot (-2i+1)}=e^{-2i^2+i}=e^{2+i}=e^2 \cdot e^i$$

Convert Euler form into Parametric form:

Since $e^{i\theta}=\cos(\theta)+i \sin(\theta)$:

We see that e^i can be converted to:

$$e^i=\cos(1)+i \sin(1)$$

Since $\theta=1$.

Now, substitute $e^i$ into $e^2 \cdot e^i$, and we are done.

Answer 2

The answer in the OP is correct, since

$e^{iQ_{-1}(i)}=e^{2+i}=e^{2}\cos(1)+ie^{2}\sin(1)$