Question

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

Answer 1

Hint: 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.