Question

Find the square root of$x + i\sqrt {{x^4} + {x^2} + 1}$where$i = \sqrt { - 1}$.

Answer 1

The square root of a number is raising the number to the power of 0.5, such as${a^{\left( {\dfrac{1}{2}} \right)}}$. This can also be denoted as $\sqrt a$ and is abbreviated as the square root of$a$. A complex number is a combination of real and imaginary parts and is denoted as$z = a + ib$. To find the square root of a complex number, equate the given equation with the standard complex number $z = a + ib$ where $a$ and $b$ are real part and imaginary part, respectively, $i$being an imaginary unit that is mathematically defined as $i = \sqrt { - 1}$. Then, square both sides of the resulting equation and compare the real and imaginary parts of the equations to determine the coefficients.

Complete step by step answer:
Let, $\sqrt {x + i\sqrt {{x^4} + {x^2} + 1} } = a + ib$
Squaring both the sides we get,

$$\sqrt {x + i\sqrt {{x^4} + {x^2} + 1} } = a + ib \\\ {\left( {\sqrt {x + i\sqrt {{x^4} + {x^2} + 1} } } \right)^2} = {\left( {a + ib} \right)^2} \\\$$

Here in the first term we can cancel square and the square root and for the second term we have to apply ${(a+b)^2}$ formula, then we will get

$$x + i\sqrt {{x^4} + {x^2} + 1} = {a^2} + {\left( {ib} \right)^2} + 2iab \\\ x + i\sqrt {{x^4} + {x^2} + 1} = {a^2} + \left( {{{\left( {\sqrt { - 1} } \right)}^2}{b^2}} \right) + i2ab \\\ x + i\sqrt {{x^4} + {x^2} + 1} = \left( {{a^2} - {b^2}} \right) + i\left( {2ab} \right) \\\$$

Now, comparing both sides of the real & imaginary parts, we get
$({a^2} - {b^2}) = x$ -- (i) And, $2ab = \sqrt {{x^4} + {x^2} + 1}$ --(ii)
Squaring above equations (i) and (ii), equations can be written as
${({a^2} - {b^2})^2} = {x^2}$ And, $4{a^2}{b^2} = {x^4} + {x^2} + 1$
We can write ${({a^2} + {b^2})^2}$ as:

$${({a^2} + {b^2})^2} = {({a^2} - {b^2})^2} + 4{a^2}{b^2} \\\ = \left( {{x^2}} \right) + \left( {{x^4} + {x^2} + 1} \right) \\\ = {x^4} + 2{x^2} + 1 \\\ = {({x^2} + 1)^2} \\\ \sqrt {{{\left( {{a^2} + {b^2}} \right)}^2}} = \sqrt {{{\left( {{x^2} + 1} \right)}^2}} \\\ \left( {{a^2} + {b^2}} \right) = \left( {{x^2} + 1} \right) \\\$$

We get ${a^2} + {b^2} = {x^2} + 1$-- (iii)
Now, solving the equations by adding (i) and (iii) to find the real term $a$of the equation we have,

$${a^2} - {b^2} + {a^2} + {b^2} = {x^2} + x + 1 \\\ 2{a^2} = {x^2} + x + 1 \\\ {a^2} = \dfrac{{{x^2} + x + 1}}{2} \\\$$

Or, $a = \sqrt {\dfrac{{{x^2} + x + 1}}{2}}$
Now for the imaginary term$b$, subtract equation (i) from (iii) we get:

$${a^2} + {b^2} - {a^2} + {b^2} = {x^2} + 1 - x \\\ 2{b^2} = {x^2} - x + 1 \\\ {b^2} = \dfrac{{{x^2} - x + 1}}{2} \\\ b = \sqrt {\dfrac{{{x^2} - x + 1}}{2}} \\\$$

Hence we have, \sqrt {x - i\sqrt {{x^4} + {x^2} + 1} } $$$$ = \sqrt {\dfrac{{{x^2} + x + 1}}{2}} $$$$ + i\sqrt {\dfrac{{{x^2} - x + 1}}{2}}

Note: In general, when we have to find an imaginary term in the given equation, we need to equate the equation with the complex equation$z = a + ib$, where $a$is the real term and $b$imaginary term and then carryout the general process of squaring, comparing and solving the resulting equations. Complex equations generally denote imaginary parts of the complex equations.