Answer
$x=0.87708+0.9492i$
$x=−0.6317+1.1275i$
$x=−1.2675−0.2524i$
$x=−0.1516−1.2835i$
$x=1.1738−0.5408i$
Work Step by Step
Step 1: Rewrite the equation in the form of $x^5=w$.
$x^5+2+3i=0$
Substracting both sides with $2+3i$,
$x^5=\underbrace{-2-3i}_{w}$
Step 2: Write the complex number $w$ in a trigonometric form.
Find the modulus of $w$:
$|w|=\sqrt{(-2)^2+(-3)^2}=\sqrt{4+9}=\sqrt{13}$
Find the argument $\angle$ of $w$:
Notice that $Re(w)=-2<0$ and $Im(w)=-3<0$, then $w$ is in quadrant IV, that is $\frac{3\pi}{2}\leq \theta\leq 2\pi$.
The value of $\theta$ is determined by solving $\tan\theta =\frac{-3}{-2}$.
Since $0<\tan^{-1}\left(\frac{-3}{-2}\right)<\frac{\pi}{2}$ and $\tan(\theta)=\tan(\theta+\pi)$, it must be that $\theta=\tan^{-1}\left(\frac{-3}{-2}\right)+\pi$.
So, $w=\sqrt{13}(\cos\left(\tan^{-1}\left(\frac{-3}{-2}\right)+\pi)\right)+i\sin\left(\tan^{-1}\left(\frac{-3}{-2}\right)+\pi)\right))$.
Step 3: Find the fifth roots of $w$ by using DeMoivre's Theorem.
$x=\sqrt[5]{w}$
$x=\sqrt[5]{r}\left(\cos\left(\frac{\left(\tan^{-1}\left(\frac{-3}{-2}\right)+\pi)\right)+2k\pi}{5}\right)+i\sin\left(\frac{\left(\tan^{-1}\left(\frac{-3}{-2}\right)+\pi)\right)+2k\pi}{5}\right)\right)$
$x=\sqrt[5]{\sqrt{13}}\left(\cos\left(\frac{\tan^{-1}\left(\frac{-3}{-2}\right)+(2k+1)\pi}{5}\right)+i\sin\left(\frac{\tan^{-1}\left(\frac{-3}{-2}\right)+(2k+1)\pi}{5}\right)\right)$
$x=\sqrt[10]{13}\cos\left(\frac{\tan^{-1}\left(\frac{-3}{-2}\right)+(2k+1)\pi}{5}\right)+i\sqrt[10]{13}\cos\left(\frac{\tan^{-1}\left(\frac{-3}{-2}\right)+(2k+1)\pi}{5}\right)$ for $k=0,1,\ldots,5$.
Now, use a calculator to finda the values of $x$:
For $k=0$, $x=0.87708+0.9492i$
For $k=1$, $x=−0.6317+1.1275i$
For $k=2$, $x=−1.2675−0.2524i$
For $k=3$, $x=−0.1516−1.2835i$
For $k=4$, $x=1.1738−0.5408i$
