Question

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

Answer 1

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.