Answer
$1.1+0.2i$ ; $-0.2+1.1i$ ; $-1.1-0.2i$ ; $0.2-1.1i$
Work Step by Step
Roots of a complex number represented in polar form can be found by the application of DeMoivre’s theorem.
DeMoivre’s theorem for finding complex roots states that every complex number has two distinct complex square roots, three distinct complex cube roots, four distinct complex fourth roots, and so on. Each root has the same modulus.
$\begin{align}
& r=z_{k}^{n} \\
& {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi k}{n} \right)+i\,\sin \left( \frac{\theta +2\pi k}{n} \right) \right] \\
\end{align}$
where $k$ is the number of distinct $n\text{th}$ root and $\theta $ is in radians.
There are exactly four fourth roots of $1+i$ as per DeMoivre’s theorem.
Convert the above expression into polar form as follows:
$\begin{align}
& 1+i=r\left( \cos \,\theta +i\,\sin \,\theta \right) \\
& =\sqrt{2}\left( \cos \left( \frac{\pi }{4} \right)+i\,\sin \left( \frac{\pi }{4} \right) \right)
\end{align}$
Apply DeMoivre’stheorem to find all fourth roots of $1+i$.
$\begin{align}
& {{z}_{k}}=\sqrt[4]{\sqrt{2}}\left[ \cos \left( \frac{\frac{\pi }{4}+2\pi k}{4} \right)+i\,\sin \left( \frac{\frac{\pi }{4}+2\pi k}{4} \right) \right] \\
& =\sqrt[8]{2}\left[ \cos \left( \frac{\frac{\pi }{4}+2\pi k}{4} \right)+i\,\sin \left( \frac{\frac{\pi }{4}+2\pi k}{4} \right) \right]
\end{align}$
Here, $k=0,1,2,3$.
Insert the above values of $k$ to find four distinct fourth roots of the provided number.
For $k=0$,
$\begin{align}
& {{z}_{0}}=\sqrt[8]{2}\left[ \cos \left( \frac{\frac{\pi }{4}+2\pi \left( 0 \right)}{4} \right)+i\,\sin \left( \frac{\frac{\pi }{4}+2\pi \left( 0 \right)}{4} \right) \right] \\
& =\sqrt[8]{2}\left[ \cos \left( \frac{\pi }{16} \right)+i\,\sin \left( \frac{\pi }{16} \right) \right]
\end{align}$
Substitute the values of $\cos \left( \frac{\pi }{16} \right)$ and $\sin \left( \frac{\pi }{16} \right)$:
$\begin{align}
& {{z}_{0}}=\sqrt[8]{2}\left[ 0.9808+i0.1951 \right] \\
& =1.0696+i0.2127 \\
& \approx 1.1+0.2i
\end{align}$
For $k=1$,
$\begin{align}
& {{z}_{1}}=\sqrt[8]{2}\left[ \cos \left( \frac{\frac{\pi }{4}+2\pi \left( 1 \right)}{4} \right)+i\,\sin \left( \frac{\frac{\pi }{4}+2\pi \left( 1 \right)}{4} \right) \right] \\
& =\sqrt[8]{2}\left[ \cos \left( \frac{9\pi }{16} \right)+i\,\sin \left( \frac{9\pi }{16} \right) \right]
\end{align}$
Substitute the values of $\cos \left( \frac{9\pi }{16} \right)$ and $\sin \left( \frac{9\pi }{16} \right)$:
$\begin{align}
& {{z}_{1}}=\sqrt[8]{2}\left[ -0.1951+i\left( 0.9808 \right) \right] \\
& =-0.2127+i1.0696 \\
& \approx -0.2+1.1i
\end{align}$
For $k=2$,
$\begin{align}
& {{z}_{2}}=\sqrt[8]{2}\left[ \cos \left( \frac{\frac{\pi }{4}+2\pi \left( 2 \right)}{4} \right)+i\,\sin \left( \frac{\frac{\pi }{4}+2\pi \left( 2 \right)}{4} \right) \right] \\
& =\sqrt[8]{2}\left[ \cos \left( \frac{17\pi }{16} \right)+i\,\sin \left( \frac{17\pi }{16} \right) \right]
\end{align}$
Substitute the values of $\cos \left( \frac{17\pi }{16} \right)$ and $\sin \left( \frac{17\pi }{16} \right)$:
$\begin{align}
& {{z}_{2}}=\sqrt[8]{2}\left[ -0.9808+i\left( -0.1951 \right) \right] \\
& =-1.0696-0.2127i \\
& \approx -1.1-0.2i
\end{align}$
For $k=3$,
$\begin{align}
& {{z}_{3}}=\sqrt[8]{2}\left[ \cos \left( \frac{\frac{\pi }{4}+2\pi \left( 3 \right)}{4} \right)+i\,\sin \left( \frac{\frac{\pi }{4}+2\pi \left( 3 \right)}{4} \right) \right] \\
& =\sqrt[8]{2}\left[ \cos \left( \frac{25\pi }{16} \right)+i\,\sin \left( \frac{25\pi }{16} \right) \right]
\end{align}$
Substitute the values of $\cos \left( \frac{25\pi }{16} \right)$ and $\sin \left( \frac{25\pi }{16} \right)$:
$\begin{align}
& {{z}_{3}}=\sqrt[8]{2}\left[ 0.1951+i\left( -0.9808 \right) \right] \\
& =0.2127-i1.0696 \\
& \approx 0.2-1.1i
\end{align}$
