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**Hint**: A complex number is a combination of real part and imaginary part. As per the complex coordinate system, $ x $ denotes the real axis and $ y $ represents the imaginary axis. The imaginary part consists of $ i $ called as imaginary unit number denotes square root of $ - 1 $ , i.e. $ i = \sqrt { - 1} $ . In the complex number of the form $ a + ib $ , $ a $ denotes the real part of the complex number and $ b $ denotes the imaginary part of the complex number. The magnitude of a complex number of the form $ a + ib $ is given as $ \sqrt {{a^2} + {b^2}} $ . Understanding of complex numbers is necessary in order to solve this type of question.

**:**

__Complete step-by-step answer__Let us assume that the square root of the given complex number is $ a + ib $ .

So, this means $ a + ib = \sqrt { - 16 + 30i} $

Now, we should square the equation on both sides.

$\Rightarrow {\left( {a + ib} \right)^2} = {\left( {\sqrt { - 16 + 30i} } \right)^2} $

We know that $ {i^2} = - 1 $ .

On simplifying the equation using the formula $ {\left( {x + y} \right)^2} = {x^2} + 2xy + {y^2} $ , we get,

$\Rightarrow {a^2} - {b^2} + i2ab = - 16 + 30i $

Let us compare both sides of the equation and equate the imaginary parts of both sides as well as the real parts present on both sides of the equation.

$ {a^2} - {b^2} = - 16 $ and $ i2ab = 30 $

$ 2ab = 30 $ implies that $ ab = 15 $

It is known that $ {\left| z \right|^2} = \left| {{z^2}} \right| $ .

Let us assume that $ - 16 + 30i = {z^2} $ , this implies that $ a + ib = z $ .

But we already know that $ {\left| z \right|^2} = \left| {{z^2}} \right| $

Let us substitute $ {z^2} = - 16 + 30i $ and $ z = a + ib $ in the equation $ {\left| z \right|^2} = \left| {{z^2}} \right| $ .

After substitution, we get, $ {\left| {a + ib} \right|^2} = \left| { - 16 + 30i} \right| $ .

The magnitude of $ a + ib $ is $ \left| {a + ib} \right| = \sqrt {{a^2} + {b^2}} $ .

So,

$ \Rightarrow {\left( {\sqrt {{a^2} + {b^2}} } \right)^2} = \left| { - 16 + 30i} \right| $ . Also $ \left| { - 16 + 30i} \right| = \sqrt {{{\left( { - 16} \right)}^2} + {{\left( {30} \right)}^2}} $

Now, $ {\left( {\sqrt {{a^2} + {b^2}} } \right)^2} = \sqrt {{{\left( { - 16} \right)}^2} + {{\left( {30} \right)}^2}} $

$

\Rightarrow {a^2} + {b^2} = \sqrt {256 + 900} \\

\Rightarrow {a^2} + {b^2} = \sqrt {1156} \\

\Rightarrow {a^2} + {b^2} = 34

$

$

ab = 15\\

a = \dfrac{{15}}{b}

$

Now, we can substitute in the equation $ {a^2} + {b^2} = 34 $ .

After substituting, we get, $ {\left( {\dfrac{{15}}{b}} \right)^2} + {b^2} = 34 $

$

\Rightarrow \dfrac{{225}}{{{b^2}}} + {b^2} = 34\\

\dfrac{{225 + {b^2} \cdot {b^2}}}{{{b^2}}} = 34\\

\dfrac{{225 + {b^4}}}{{{b^2}}} = 34

$

After multiplying $ {b^2} $ on both sides of the equation, we get $ 225 + {b^4} = 34{b^2} $

On rearranging the terms, we get, $ {b^4} - 34{b^2} + 225 = 0 $

It is known that $ 25 + 9 = 34 $ and $ 25 \times 9 = 225 $ .

So, $ {b^4} - 34{b^2} + 225 = 0 $ can be written as $ {b^4} - 25{b^2} - 9{b^2} + 225 = 0 $ .

Now, let us take $ {b^2} $ and 9 common from the equation $ {b^4} - 25{b^2} - 9{b^2} + 225 = 0 $

$ {b^2}\left( {{b^2} - 25} \right) - 9\left( {{b^2} - 25} \right) = 0\\

\left( {{b^2} - 25} \right)\left( {{b^2} - 9} \right) = 0 $

This implies, $ {b^2} - 9 = 0 $ and $ {b^2} - 25 = 0 $

So, $ b = \pm 3 $ and $ b = \pm 5 $

If b = 3, then $ a = 5 $ as $ a \cdot b = 15 $

If $ b = - 3 $ , then $ a = - 5 $ as $ a \cdot b = 15 $

If $ b = 5 $ , then $ a = 3 $ as $ a \cdot b = 15 $

If $ b = - 5 $ , then $ a = - 3 $ as $ a \cdot b = 15 $

But $ {a^2} - {b^2} = - 16 $ . So, the values of $ a $ and $ b $ should also satisfy the equation $ {a^2} - {b^2} = - 16 $ .

Now, let us substitute a = 3 and $ b = 5 $ to check whether $ {a^2} - {b^2} $ is $ - 16 $ .

$

\Rightarrow {a^2} - {b^2} = {3^2} - {5^2}\\

= 9 - 25\\

= - 16

$

So, a = 3 and $ b = 5 $ satisfies the equation $ {a^2} - {b^2} = - 16 $ .

Now, let us substitute a = - 3 and $ b = - 5 $ to check whether $ {a^2} - {b^2} $ is $ - 16 $

$

\Rightarrow {a^2} - {b^2} = {\left( { - 3} \right)^2} - {\left( { - 5} \right)^2}\\

= 9 - 25\\

= - 16

$

So, a = - 3 and $ b = - 5 $ satisfies the equation $ {a^2} - {b^2} = - 16 $ .

Now, let us substitute a = - 5 and $ b = - 3 $ to check whether $ {a^2} - {b^2} $ is $ - 16 $ .

$

\Rightarrow {a^2} - {b^2} = {\left( { - 5} \right)^2} - {\left( { - 3} \right)^2}\\

= 25 - 9\\

= 16

$

So, a = - 5 and $ b = - 3 $ doesn’t satisfy the equation $ {a^2} - {b^2} = - 16 $ .

Now, let us substitute $ a = 5 $ and $ b = 3 $ to check whether $ {a^2} - {b^2} $ is $ - 16 $

$

\Rightarrow {a^2} - {b^2} = {5^2} - {3^2}\\

= 25 - 9\\

= 16

$

So, $ a = 5 $ and $ b = 3 $ doesn’t satisfy the equation $ {a^2} - {b^2} = - 16 $ .

$ a = 3 $ , $ b = 5 $ and $ a = - 3 $ , $ b = - 5 $ .

The required complex numbers $ a + ib $ are $ - 3 - 5i $ and $ 3 + 5i $ .

Therefore, the square root of the complex number $ - 16 + 30i $ are $ - 3 - 5i $ and $ 3 + 5i $ .

**So, the correct answer is “ $ - 3 - 5i $ and $ 3 + 5i $ ”.**

**Note**: In this type of question, students use the formula for finding the magnitude of complex numbers properly without making any mistakes. Also, note that when the square of a variable is given as a constant, then the values of the variable are positive and negative of the square root of the constant. This can be represented mathematically as, if $ {x^2} = a $ then $ x = \pm a $ .

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