Answer
$7\left( \cos 25{}^\circ +i\sin 25{}^\circ \right),7\left( \cos 205{}^\circ +i\sin 205{}^\circ \right)$
Work Step by Step
Method for finding complex roots:
To find $n$ distinct complex roots for any complex number $z=r\left( \cos \theta +i\sin \theta \right)$, if $z\ne 0$, in radians we use the formula given below:
${{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +360{}^\circ \cdot k}{n} \right)+i\sin \left( \frac{\theta +360{}^\circ \cdot k}{n} \right) \right]$
Where $k=0,1,2,3,....,n-1$.
So, the square roots of $49\left( \cos 50{}^\circ +i\sin 50{}^\circ \right)$ are
${{z}_{k}}=\sqrt[2]{49}\left[ \cos \left( \frac{50{}^\circ +360{}^\circ \cdot k}{2} \right)+i\sin \left( \frac{50{}^\circ +360{}^\circ \cdot k}{2} \right) \right],k=0,1$
Therefore, we can find the two complex square roots in the following manner:
$\begin{align}
& {{z}_{0}}=\sqrt[2]{49}\left[ \cos \left( \frac{50{}^\circ +360{}^\circ \cdot 0}{2} \right)+i\sin \left( \frac{50{}^\circ +360{}^\circ \cdot 0}{2} \right) \right] \\
& =7\left( \cos \frac{50{}^\circ }{2}+i\sin \frac{50{}^\circ }{2} \right) \\
& =7\left( \cos 25{}^\circ +i\sin 25{}^\circ \right)
\end{align}$
$\begin{align}
& {{z}_{1}}=\sqrt[2]{49}\left[ \cos \left( \frac{50{}^\circ +360{}^\circ \cdot 1}{2} \right)+i\sin \left( \frac{50{}^\circ +360{}^\circ \cdot 1}{2} \right) \right] \\
& =7\left( \cos \frac{410{}^\circ }{2}+i\sin \frac{410{}^\circ }{2} \right) \\
& =7\left( \cos 205{}^\circ +i\sin 205{}^\circ \right)
\end{align}$
So, the square roots are $7\left( \cos 25{}^\circ +i\sin 25{}^\circ \right)\text{ and }7\left( \cos 205{}^\circ +i\sin 205{}^\circ \right)$.
