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# Find the square root of $- 16 + 30i$ .

Last updated date: 23rd Feb 2024
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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.

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 $x = \pm a$ .