How do you find the square root of an imaginary number of the form \[a + bi\] ?

Hint: An imaginary number is the square root of a negative real number i.e., the square root of a number is a second number that, when multiplied by itself, equals the first number and to find the square root of an imaginary number of the form \[a + bi\] , we need to consider any form of it as we took; \[a + bi = {\left( {c + di} \right)^2}\] , such that solve the terms by applying quadratic formula and find the square root of an imaginary number to get the required expression.
Formula used:
 \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]
 \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

Complete step-by-step answer:
We need to find the square root of an imaginary number of the form \[a + bi\] .
If, suppose
 \[a + bi = {\left( {c + di} \right)^2}\]
Here, we need to solve for c and d; as, we know that \[{\left( {c + di} \right)^2}\] is of the form \[{\left( {a + b} \right)^2}\] i.e., \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\] hence, applying this we have:
 \[{\left( {c + di} \right)^2} = {c^2} + 2cdi + {d^2}{i^2}\]
 \[ = \left( {{c^2} - {d^2}} \right) + \left( {2cd} \right)i\]
Now, we need to solve for
 \[\left( {{c^2} - {d^2}} \right) = a\] ……………… 1
 \[2cd = b\] …………………. 2
From the second equation of these, we need to solve for \[{d^2}\] i.e.,
 \[ \Rightarrow d = \dfrac{b}{{2c}}\]
So, squaring both sides we get:
 \[{d^2} = \dfrac{{{b^2}}}{{{2^2}{c^2}}}\]
 \[ \Rightarrow {d^2} = \dfrac{{{b^2}}}{{4{c^2}}}\] ……………………… 3
Now, substitute the value of \[{d^2}\] from equation 3 in equation 1 we get:
 \[\left( {{c^2} - {d^2}} \right) = a\]
 \[ \Rightarrow \left( {{c^2} - \dfrac{{{b^2}}}{{4{c^2}}}} \right) = a\] …………………. 4
Multiply equation 4 by \[4{c^2}\] as:
 \[4{c^2}{c^2} - \dfrac{{{b^2}}}{{4{c^2}}}4{c^2} = 4a{c^2}\]
As, the numerator and denominator term \[4{c^2}\] are same which implies to one, hence we get:
 \[ \Rightarrow 4{\left( {{c^2}} \right)^2} - {b^2} = 4a{c^2}\]
Now, subtract \[4a{c^2}\] from both sides of the obtained equation as:
 \[4{\left( {{c^2}} \right)^2} - {b^2} - 4a{c^2} = 4a{c^2} - 4a{c^2}\]
Simplifying we get:
 \[ \Rightarrow 4{\left( {{c^2}} \right)^2} - {b^2} - 4a{c^2} = 0\] ………………….. 5
Now, from the quadratic formula, we find the value of \[{c^2}\] as we know that quadratic formula is given as: \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Now, substitute the values of a, b and c from equation 5, which is of the form \[a{x^2} + bx + c = 0\] ;according to this formula we have \[a = 4\] , \[b = - 4a\] and \[c = - {b^2}\] :
 \[{c^2} = \dfrac{{4a \pm \sqrt {{{\left( {4a} \right)}^2} - 4\left( 4 \right)\left( { - {b^2}} \right)} }}{8}\]
 \[{c^2} = \dfrac{{4a \pm \sqrt {{{\left( {4a} \right)}^2} + 16{b^2}} }}{8}\]
Simplifying the terms, we get:
 \[ \Rightarrow {c^2} = \dfrac{{a \pm \sqrt {{a^2} + {b^2}} }}{2}\]
For c to be Real valued we require \[{c^2} \geqslant 0\] , hence it should be + i.e.,
 \[{c^2} = \dfrac{{a + \sqrt {{a^2} + {b^2}} }}{2}\]
Hence, the value of c is:
 \[ \Rightarrow c = \sqrt {\dfrac{{\sqrt {{a^2} + {b^2}} + a}}{2}} \] …………………….. 6
Then, from equation 1 we get the value of d as:
 \[\left( {{c^2} - {d^2}} \right) = a\]
 \[ \Rightarrow d = \pm \sqrt {{c^2} - a} \] …………………….. 7
Now, substitute the value of \[{c^2}\] from equation 6 in equation 7 we get:
 \[d = \pm \sqrt {\dfrac{{\sqrt {{a^2} + {b^2}} + a}}{2} - a} \]
 \[ \Rightarrow d = \pm \sqrt {\dfrac{{\sqrt {{a^2} + {b^2}} - a}}{2}} \]
Now, since \[2cd = b\] i.e., equation 2 we have:
If \[b > 0\] then c and d must have the same signs.
If \[b < 0\] then c and d must have opposite signs.
If \[b = 0\] then \[d = 0\] .
If \[b \ne 0\] then we can use \[\dfrac{b}{{\left| b \right|}}\] as a multiplier to match the signs as we require to find that the square roots of \[a + bi\] are:
 \[ \pm \left( {\left( {\sqrt {\dfrac{{\sqrt {{a^2} + {b^2}} + a}}{2}} } \right) + \left( {\dfrac{b}{{\left| b \right|}}\sqrt {\dfrac{{\sqrt {{a^2} + {b^2}} - a}}{2}} } \right)i} \right)\]

Note: To find the square root of \[a + bi\] we need to note that:
For positive Real numbers x, the principal square root of x is the positive one, which is the one we mean when we write \[\sqrt x \] . This is also the square root that people commonly mean when they say "the square root of 2" and suchlike, neglecting the fact that 2 has two square roots.
For negative Real numbers x, then by convention and definition, \[\sqrt x = i\sqrt { - x} \] , i.e., the principal square root is the one with a positive coefficient of i.