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**Hint**: We first try to form the imaginary numbers for $-59$ using the identities ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$. Then we get the root value in the form of both positive and negative values and imaginary numbers. We also find the decimal value for $\sqrt{59}$.

**:**

__Complete step-by-step answer__The square root value of the negative value gives imaginary values.

Therefore, we use the know identity values and relations for imaginary $i=\sqrt{-1}$ where ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$.

Therefore, $-59=\left( -1 \right)\times 59=59{{i}^{2}}$.

Now taking the root value we get $\sqrt{-59}=\sqrt{59{{i}^{2}}}=\pm i\sqrt{59}$.

We can also find the root value in decimal for 59 as 59 is a prime number.

We take 2 digits as a set from the right end and complete the division. For decimal form we take the set from the right side of the decimal.

\[\begin{align}

& 7 \\

& 7\left| \!{\overline {\,

\begin{align}

& \overline{59}.\overline{00}\overline{00} \\

& \underline{49} \\

& 10.00 \\

\end{align} \,}} \right. \\

\end{align}\]

Now we have to enter the decimal part. We keep doing the breaking in the set form till 3-digit place after decimal.

\[\begin{align}

& 7.67 \\

& 146\left| \!{\overline {\,

\begin{align}

& 1000\overline{00} \\

& \underline{876} \\

& 12400 \\

\end{align} \,}} \right. \\

& 1527\left| \!{\overline {\,

\begin{align}

& 12400 \\

& \underline{10689} \\

& 1711 \\

\end{align} \,}} \right. \\

\end{align}\]

So, $\sqrt{-59}=\pm i\sqrt{59}=\pm 7.67i$

Therefore, the square root of $-59$ is $\pm i\sqrt{59}=\pm 7.67i$.

**So, the correct answer is “$\pm 7.67i$”.**

**Note**: The long-division method and arranging the set of 2 digits is different for integer and decimal. But taking double for the next division and putting a particular number is the same process for both of them. Since 3 is a non-perfect square number, we will find the value of root 3 using the long division method as shown above.

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