Question

If $g\left( x \right)={{x}^{4}}-{{x}^{3}}+{{x}^{2}}+3x-5$, then find $g\left( 2+3i \right)$.

Answer 1

Hint: We will first find all the values like ${{\left( 2+3i \right)}^{4}},{{\left( 2+3i \right)}^{3}},{{\left( 2+3i \right)}^{2}},3\left( 2+3i \right)$ and then substitute them in the function $g\left( x \right)$ and get the calculated value. That would be our final answer.

Complete step-by-step answer:

Let, $x=2+3i$. Let us calculate ${{x}^{2}}$ first.
Therefore, ${{x}^{2}}={{\left( 2+3i \right)}^{2}}$.
Expanding using the formula, ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$, we get,

& {{x}^{2}}={{2}^{2}}+2\times 2\times 3i+{{\left( 3i \right)}^{2}} \\\ & {{x}^{2}}=4+12i+9{{i}^{2}} \\\ \end{aligned}$$ As, $$i=\sqrt{-1}$$, we get $${{i}^{2}}=-1$$. Thus, $${{x}^{2}}=4+12i+9\left( -1 \right)$$ $${{x}^{2}}=4+12i-9$$ $${{x}^{2}}=-5+12i-(i)$$ Now, let us calculate, $${{x}^{3}}$$. $${{x}^{3}}={{\left( 2+3i \right)}^{3}}$$ Thus, $${{x}^{3}}={{\left( 2+3i \right)}^{2}}\left( 2+3i \right)$$ We know, $${{\left( 2+3i \right)}^{2}}=\left( -5+12i \right)$$ Thus, $${{x}^{3}}=\left( -5+12i \right)\left( 2+3i \right)$$ Multiplying we get, $$\begin{aligned} & {{x}^{3}}=\left( -5\times 2 \right)+\left( -5\times 3i \right)+\left( 12i\times 2 \right)+\left( 12i\times 3i \right) \\\ & {{x}^{3}}=-10-15i+24i+36{{i}^{2}} \\\ & {{x}^{3}}=-10+9i+36{{i}^{2}} \\\ \end{aligned}$$ Putting, $${{i}^{2}}=-1$$ we get, $$\begin{aligned} & {{x}^{3}}=-10+9i+36\left( -1 \right) \\\ & {{x}^{3}}=-10+9i-36 \\\ & {{x}^{3}}=\left( -46+9i \right)-(ii) \\\ \end{aligned}$$ Now let us calculate $${{x}^{4}}$$. We know, $${{x}^{2}}=-5+12i$$ from (i). Squaring on both sides we get, $${{\left( {{x}^{2}} \right)}^{2}}={{\left( -5+12i \right)}^{2}}$$ Expanding using, $${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$$, we get, $$\begin{aligned} & {{x}^{4}}={{\left( -5 \right)}^{2}}+2\times \left( -5 \right)\left( 12i \right)+{{\left( 12i \right)}^{2}} \\\ & {{x}^{4}}=25-120i+144{{i}^{2}} \\\ \end{aligned}$$ Putting $${{i}^{2}}=-1$$, we get, $$\begin{aligned} & {{x}^{4}}=25-120i+144\left( -1 \right) \\\ & {{x}^{4}}=25-120i-144 \\\ & {{x}^{4}}=-119-120i-(iii) \\\ \end{aligned}$$ Now that we know all the values in $$g\left( x \right)$$, let’s put them in. $$g\left( x \right)={{x}^{4}}-{{x}^{3}}+{{x}^{2}}+3x-5$$ $$\Rightarrow g\left( 2+3i \right)=\left( -119-120i \right)-\left( -46+9i \right)+\left( -5+12i \right)+3\left( 2+3i \right)-5$$ Multiplying and simplifying, $$\begin{aligned} & \Rightarrow g\left( 2+3i \right)=-119+120i+46-9i-5+12i+6+9i-5 \\\ & \Rightarrow g\left( 2+3i \right)=-77-108i \\\ \end{aligned}$$ Thus, $$g\left( 2+3i \right)=-77-108i$$. Note: Be careful solving these types of huge equations. Do not put the complex number directly into the equation and start expanding there itself. This will just reduce your steps but it invites a lot of confusion. The equation becomes really huge and you will commit a mistake by messing up the signs or missing some terms in the middle. So a better practice is that we calculate every term separately and then put all the terms in the equation and calculate.