Question

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

Answer 1

Hint: 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 values\[x = 1\]\[\& \]\[x = - 1\] 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\[( - 8 - 6i)\] is\[1 - 3i\]\[\& \]\[ - 1 + 3i\].
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.