Question

Find the square root of $6 + 8i$

Answer 1

Hint : Complex number contains the real part and an imaginary part and it is denoted by “Z”. It is expressed as $z = a + ib$ where “a” is the real part and “b” is the imaginary part of the complex number. Here, we will Compare real and imaginary parts and simplify for the resultant values.

Complete step-by-step answer :
Given expression is –
$z = \sqrt {6 + 8i}$
Place, $z = a + ib$ in the above equation –
$\Rightarrow a + ib = \sqrt {6 + 8i}$
Take square on both the sides of the equations-
$\Rightarrow {\left( {a + ib} \right)^2} = {\left( {\sqrt {6 + 8i} } \right)^2}$
Expand the brackets by using the sum of two terms whole square identity - ${(a + b)^2} = {a^2} + 2ab + {b^2}$ . Also, square and square root cancel each other on the right hand side of the equation.
$\Rightarrow {a^2} + 2abi + {b^2}{i^2} = 6 + 8i$
We know that ${i^2} = ( - 1)$ place in the above equation –
$\Rightarrow {a^2} + 2abi + {b^2}( - 1) = 6 + 8i$
When you multiply one positive term with negative term the resultant value is negative.
$\Rightarrow {a^2} + 2abi - {b^2} = 6 + 8i$
The above equation can be re-written as –
$\Rightarrow {a^2} - {b^2} + 2abi = 6 + 8i$
Compare the real part a9nd the imaginary part on both sides of the equation.
$\Rightarrow {a^2} - {b^2} = 6$ .... (A) and
$2ab = (8)$
Take out common multiple. Common multiple on both the sides of the equation cancels each other.
$\Rightarrow ab = 4$ .... (B)
We have two equations and two variables-
Take equation (B)
$ab = 4 \\\ \Rightarrow b = \dfrac{4}{a} \;$
Place the above value in equation (A)
$\Rightarrow {a^2} - {\left( {\dfrac{4}{a}} \right)^2} = 6$
Simplify –
$\Rightarrow {a^2} - \dfrac{{16}}{{{a^2}}} = 6$
Take LCM for the above expression -
$\Rightarrow \dfrac{{{a^4} - 16}}{{{a^2}}} = 6$
Do-cross multiplication – also place
$\Rightarrow {a^4} - 16 = 6{a^2}$
Take all the terms on one side of the equation-
$\Rightarrow {a^4} - 6{a^2} - 16 = 0$
The above equation can be re-written as –
$\Rightarrow \underline {{a^4} - 8{a^2}} + 2\underline {{a^2} - 16} = 0$
Take the common multiples –

$$\Rightarrow {a^2}({a^2} - 8) + 2({a^2} - 8) = 0 \\\ \Rightarrow ({a^2} + 2)({a^2} - 8) = 0 \;$$

We get,
$\Rightarrow {a^2} = 8{\text{ or }}{a^2} = - 2$
Since, a is the real value, we get $a = \pm 2\sqrt 2$
Place in equation (B)
$\Rightarrow ab = 4$
When $a = 2\sqrt 2 \Rightarrow b = \sqrt 2$
And When $a = - 2\sqrt 2 \Rightarrow b = \sqrt 2$
Hence, the square-roots are –
$2\sqrt 2 + i\sqrt 2 {\text{ and - }}2\sqrt 2 + i\sqrt 2 {\text{ }}$
So, the correct answer is “ $2\sqrt 2 + i\sqrt 2 {\text{ and - }}2\sqrt 2 + i\sqrt 2 {\text{ }}$ ”.

Note : Simplify wisely while finding the roots of the equations. Always remember that the middle term in the perfect square should be equal to the product of the first term and the last term while applying the split of the middle term. Always keeping the signs of the terms double check. Be good in square and square root concepts.