Question

Find the value of ${{x}^{4}}-4{{x}^{3}}+4{{x}^{2}}+8x+44$ when the value of x is 3 + 2i:

Answer 1

Hint: As in the question x is a complex number so, first we will find ${{x}^{4}}$ then ${{x}^{3}}$ and then ${{x}^{2}}$. The value of x is given hence, after that we will multiply all the power of x by their respective coefficient and then add it to get our final answer.

Complete step-by-step answer:
First we will write the formulas that we are going to use.
Given x = 3 + 2i,
The formula for ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ , we are going to use this formula for calculating the value of ${{z}^{2}}$, where z can any complex number.
Another formula that we are going to use is ${{i}^{2}}=-1$ ,
So, first ${{x}^{2}}$ case:
First we will find the value of ${{x}^{2}}$ ,
Now squaring both side of the equation x = 3 + 2i we get,
$\Rightarrow {{x}^{2}}={{\left( 3+2i \right)}^{2}}$
Now we will use ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ to expand,
$\Rightarrow {{x}^{2}}=\left( 9+12i+4{{i}^{2}} \right)$
Now we know that ${{i}^{2}}=-1$ , using this we get,
$\Rightarrow {{x}^{2}}=\left( 5+12i \right)$
As we have found the value of ${{x}^{2}}$.
Now ${{x}^{3}}$ case:
Now let’s find ${{x}^{3}}$ using ${{x}^{2}}$,
$\begin{align}
  & {{x}^{3}}=x.{{x}^{2}} \\
 & \Rightarrow {{x}^{3}}=\left( 3+2i \right){{\left( 3+2i \right)}^{2}} \\
\end{align}$
Now we will use the value of ${{x}^{2}}$ from above,
$\begin{align}
  & \Rightarrow {{x}^{3}}=\left( 3+2i \right)\left( 5+12i \right) \\
 & \Rightarrow {{x}^{3}}=\left( 15+36i+10i+24{{i}^{2}} \right) \\
\end{align}$
Now we know that ${{i}^{2}}=-1$ , using this we get,
$\Rightarrow {{x}^{3}}=\left( -9+46i \right)$
As we have found the value of ${{x}^{3}}$.
Now ${{x}^{4}}$ case:
Now let’s find the value of ${{x}^{4}}$ using ${{x}^{3}}$,
$\begin{align}
  & {{x}^{4}}=x.{{x}^{3}} \\
 & \Rightarrow {{x}^{4}}=\left( 3+2i \right){{\left( 3+2i \right)}^{3}} \\
\end{align}$
Now we will use the value of ${{x}^{3}}$ from above,
$\begin{align}
  & \Rightarrow {{x}^{4}}=\left( 3+2i \right)\left( -9+46i \right) \\
 & \Rightarrow {{x}^{4}}=\left( -27+138i-18i+92{{i}^{2}} \right) \\
\end{align}$
Now we know that ${{i}^{2}}=-1$ , using this we get,
$\Rightarrow {{x}^{4}}=\left( -119+120i \right)$
Now we have found the value of all the power of x that was needed.
Now we will just multiply the respective coefficients for different powers of x.
${{x}^{4}}-4{{x}^{3}}+4{{x}^{2}}+8x+44$
After putting the value of x, ${{x}^{2}},{{x}^{3}}$ and ${{x}^{4}}$ in the given equation ${{x}^{4}}-4{{x}^{3}}+4{{x}^{2}}+8x+44$ we get,
$\begin{align}
  & \left( -119+120i \right)-4\left( -9+46i \right)+4\left( 5+12i \right)+8\left( 3+2i \right)+44 \\
 & \Rightarrow \left( -119+36+20+24+44 \right)+\left( 120-184+48+16 \right)i \\
 & \Rightarrow \left( 5 \right)+0i \\
 & \Rightarrow 5 \\
\end{align}$
As we can see that the final answer that we got is purely real or the imaginary part is zero.
So, the value of the equation after putting x = 3 + 2i, is 5.
Hence, the answer to this question is 5.

Note: Another method to solve this question will be to directly put the value of x in the given equation and then try to solve it, but it will be a little bit complicated. And there are some things which one should know before solving this question, that the value of i is $\sqrt{-1}$ and ${{i}^{2}}=-1$, ${{i}^{3}}=-i$ and ${{i}^{4}}=1$, after that it repeats itself. So, this much is needed for the second approach.