Question

(i) Find the square root of the complex number $z=3+4i$.
(ii) Find the value of ${{x}^{3}}-{{x}^{2}}+x+46$ if $x=2+3i$.

Answer 1

Hint: We first assume a complex number as the root of the given number $z=3+4i$. The square of the variable part will give $z=3+4i$. We equate the real parts of the square to find the variables. We put the values to find the square root. In the second case we put the values in the polynomial and use the identities of ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$ to find the answer.

Complete step by step answer:
We need to find the square root of the complex number $z=3+4i$.
As there is a complex number ‘i’ present in the square value, the root will also be complex.
Let the root be $\left( a+ib \right)$. So, ${{\left( a+ib \right)}^{2}}=z=3+4i$. We know ${{i}^{2}}=-1$.
We expand the square and get ${{\left( a+ib \right)}^{2}}=\left( {{a}^{2}}-{{b}^{2}} \right)+2aib=3+4i$.
We can equate the real parts and get $\left( {{a}^{2}}-{{b}^{2}} \right)=3,2abi=4i$.
The value of ab becomes $ab=\dfrac{4}{2}=2$.
Putting the value, we get $a=\pm 2,b=\pm 1$ as $\left( {{2}^{2}}-{{1}^{2}} \right)=3$. The root is $\left( a+ib \right)=\pm \left( 2+i \right)$
So, the root of the number $z=3+4i$ is $\pm \left( 2+i \right)$.
Now we need to find the value of ${{x}^{3}}-{{x}^{2}}+x+46$ if $x=2+3i$.
We have the identities of ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$. We put the values and get
$\begin{align}
  & {{x}^{3}}-{{x}^{2}}+x+46 \\
 & ={{\left( 2+3i \right)}^{3}}-{{\left( 2+3i \right)}^{2}}+\left( 2+3i \right)+46 \\
\end{align}$
We now use the square and cube formulas to find the simplest form.
$\begin{align}
  & {{\left( 2+3i \right)}^{3}}-{{\left( 2+3i \right)}^{2}}+\left( 2+3i \right)+46 \\
 & =8+27{{i}^{3}}+36i+54{{i}^{2}}-4-9{{i}^{2}}-12i+2+3i+46 \\
\end{align}$
We place the values of ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$.
$\begin{align}
  & 8+27{{i}^{3}}+36i+54{{i}^{2}}-4-9{{i}^{2}}-12i+2+3i+46 \\
 & =52-27i+36i-54+9-12i+3i \\
 & =7 \\
\end{align}$

Therefore, the value of ${{x}^{3}}-{{x}^{2}}+x+46$ if $x=2+3i$ is 7.

Note: We can solve the equations like $\left( {{a}^{2}}-{{b}^{2}} \right)=3,ab=2$ by using the formula $b=\dfrac{2}{a}$ in the equation of $\left( {{a}^{2}}-{{b}^{2}} \right)=3$. We get quadratic equation for ${{a}^{2}}$ as ${{a}^{2}}-{{\left( \dfrac{2}{a} \right)}^{2}}=3$. We assume
${{a}^{2}}=z$. We get ${{z}^{2}}-3z-4=0\Rightarrow z=-4,1$. As z is a square value of ${{a}^{2}}=z=1$. So, $a=\pm 1$ which gives $b=\pm 2$.