Question

Root $( - 8 - 6i)$ equals
A. $1 \pm 3i$
B. $\pm (1 - 3i)$
C. $\pm (1 + 3i)$
D. $\pm (3 - i)$

Answer 1

A complex number is a number represented in the form where is an integer and is an imaginary number. When asked to discover the root of a complex number, we will first assume a general complex number as a result. Then equating that to the root of a specific complex number, solving and for the variables in the assuming complex number, and substituting them will give us the answer.

Formula used:
Some formulas that we need to know to solve this problem:
${\left( {\sqrt a } \right)^2} = a$
${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$
For an imaginary number ${i^2} = - 1$.

Complete step by step answer:
We aim to find the root of the complex number$( - 8 - 6i)$. Let us assume that the complex number $x + iy$is the square root of the given complex number$( - 8 - 6i)$ where $x\& y$are real numbers and $i$is an imaginary number.
Therefore, we can write it as $x + iy = \sqrt { - 8 - 6i}$
Now let us solve this equation for the unknown variables$x\& y$.
Consider $x + iy = \sqrt { - 8 - 6i}$
To remove the square root on the right-hand side, let us square the equation on both sides.
$\Rightarrow {(x + iy)^2} = {\left( {\sqrt { - 8 - 6i} } \right)^2}$
By using the formula ${\left( {\sqrt a } \right)^2} = a$we get
$\Rightarrow {(x + iy)^2} = ( - 8 - 6i)$
Now let us expand the term on the left-hand side by using the formula${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$.
$\Rightarrow {x^2} + 2xiy + {\left( {iy} \right)^2} = ( - 8 - 6i)$
We know that${i^2} = - 1$ by applying this to the above equation, we get
$\Rightarrow {x^2} + 2xiy - {y^2} = ( - 8 - 6i)$
On equating real terms and imaginary terms, we get
${x^2} - {y^2} = - 8$……..$(1)$
$2xy = - 6$………$(2)$
On simplifying the equation $(2)$, we get
$2xy = - 6 \Rightarrow xy = - 3$
$\Rightarrow y = \dfrac{{ - 3}}{x}$……..$(3)$
Now let’s substitute the value $y = \dfrac{{ - 3}}{x}$ in the equation $(1)$
(1)$$$$ \Rightarrow {x^2} - {\left( {\dfrac{{ - 3}}{x}} \right)^2} = - 8
On solving the above equation, we get
$\Rightarrow {x^2} - \left( {\dfrac{9}{{{x^2}}}} \right) = - 8$
On further simplification, we get
$\Rightarrow {x^4} - 9 = - 8{x^2}$
$\Rightarrow {x^4} + 8{x^2} - 9 = 0$
Now let us factorize the above equation.
$\Rightarrow {x^4} + 9{x^2} - {x^2} - 9 = 0$
$\Rightarrow {x^2}({x^2} - 1) + 9({x^2} - 1) = 0$
Let us take the term $({x^2} - 1)$commonly out.
$\Rightarrow ({x^2} - 1)({x^2} + 9) = 0$
$\Rightarrow ({x^2} - 1) = 0\& ({x^2} + 9) = 0$
From this we get,
$x = \pm 1$ $\&$ $x = \sqrt { - 9}$
Since the variable $x$is a real number, we take$x = \pm 1$ that is x = 1$$$$\& $$$$x = - 1
Substitute the valuesx = 1$$$$\& $$$$x = - 1 in the equation $(3)$, we get
When $x = 1$, we get
(3)$$$$ \Rightarrow y = \dfrac{{ - 3}}{1} = - 3
When $x = - 1$, we get
(3)$$$$ \Rightarrow y = \dfrac{{ - 3}}{{ - 1}} = 3
Thus when$x = 1$,$y = - 3$ and when$x = - 1$, $y = 3$
Therefore, the square root of the given complex number$( - 8 - 6i)$ is1 - 3i$$$$\& $$$$ - 1 + 3i.
Which can be written as$\pm (1 - 3i)$.
Now let us see the options, option (1) $1 \pm 3i$cannot be the correct answer since we got that $\pm (1 - 3i)$from our calculation.
Option (2) $\pm (1 - 3i)$is the correct answer as we got the same answer in our calculation.
Option (3) $\pm (1 + 3i)$cannot be the correct answer since we got that $\pm (1 - 3i)$from our calculation.
Option (4) $\pm (3 - i)$cannot be the correct answer since we got that $\pm (1 - 3i)$from our calculation.
Hence, option (2) $\pm (1 - 3i)$is the correct answer.

Note: An imaginary number is nothing but the square root of minus one (that is$\sqrt { - 1} = i$). Thus, the value of the imaginary$i$ is minus one (that is${i^2} = \sqrt { - 1} \times \sqrt { - 1} = - 1$). When two complex numbers are equal then their real and imaginary parts are also equal.