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Hint: It is given that $x=1+2i$. Using this, find out ${{x}^{2}}$ and ${{x}^{3}}$ and then substitute

$x,{{x}^{2}},{{x}^{3}}$ in the equation which we have to prove in the question.

In the question, we are given a complex number $x=1+2i$. We have to prove that the expression ${{x}^{3}}+7{{x}^{2}}-13x+16$ is equal to $-29$.

In the expression ${{x}^{3}}+7{{x}^{2}}-13x+16$, we can see that ${{x}^{2}}$ and ${{x}^{3}}$ are present. So, we have to find both ${{x}^{2}}$ and ${{x}^{3}}$.

In the question, it is given $x=1+2i$.

Squaring both the sides of the above equation, we can find ${{x}^{2}}$ as,

${{x}^{2}}={{\left( 1+2i \right)}^{2}}$

We have a formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.

Using this formula to find ${{x}^{2}}$, we get,

$\begin{align}

& {{x}^{2}}={{\left( 1 \right)}^{2}}+{{\left( 2i \right)}^{2}}+2\left( 1 \right)\left( 2i \right) \\

& \Rightarrow {{x}^{2}}=1+4{{i}^{2}}+4i \\

\end{align}$

In complex numbers, we have a formula ${{i}^{2}}=-1$.

Substituting ${{i}^{2}}=-1$in the above equation, we get,

\[\begin{align}

& {{x}^{2}}=1-4+4i \\

& \Rightarrow {{x}^{2}}=-3+4i...............\left( 1 \right) \\

\end{align}\]

To find ${{x}^{3}}$, we will multiply the above equation with $x=1+2i$.

\[\begin{align}

& {{x}^{2}}.x=\left( -3+4i \right)\left( 1+2i \right) \\

& \Rightarrow {{x}^{3}}=-3-6i+4i+8{{i}^{2}} \\

& \Rightarrow {{x}^{3}}=-3-2i+8{{i}^{2}} \\

\end{align}\]

Substituting ${{i}^{2}}=-1$in the above equation, we get,

\[\begin{align}

& {{x}^{3}}=-3-2i-8 \\

& \Rightarrow {{x}^{3}}=-11-2i..........\left( 2 \right) \\

\end{align}\]

Since in the question we have to prove ${{x}^{3}}+7{{x}^{2}}-13x+16=-29$, substituting ${{x}^{2}}$

from equation $\left( 1 \right)$ and ${{x}^{3}}$ from equation $\left( 2 \right)$ in

${{x}^{3}}+7{{x}^{2}}-13x+16$, we get,

\[\begin{align}

& -11-2i+7\left( -3+4i \right)-13\left( 1+2i \right)+16 \\

& \Rightarrow -11-2i-21+28i-13-26i+16 \\

& \Rightarrow -29 \\

\end{align}\]

Hence, we have proved that ${{x}^{3}}+7{{x}^{2}}-13x+16=-29$.

Note: One can also do this question by converting the complex number $x=1+2i$ to the Euler’s form i.e. in the form of $r{{e}^{i\theta }}$ or $r\left( \cos \theta +i\sin \theta \right)$ where $r$ is the modulus of the complex number and $\theta $ is the argument of the complex number. Then using the De Moivre’s theorem i.e. ${{\left( r{{e}^{i\theta }} \right)}^{n}}=\cos n\theta +i\sin n\theta $, one can find ${{x}^{2}}$ and ${{x}^{3}}$ and substituting $x,{{x}^{2}},{{x}^{3}}$ in the expression ${{x}^{3}}+7{{x}^{2}}-13x+16$. But this method will take a lot of time since the argument of the complex number is not a standard angle. So, one has to write $\sin 2\theta ,\sin 3\theta ,\cos 2\theta ,\cos 3\theta $ in terms of $\sin \theta $ and $\cos \theta $ using the trigonometric formulas. Then, one has to simplify all the terms to get the answer.

$x,{{x}^{2}},{{x}^{3}}$ in the equation which we have to prove in the question.

In the question, we are given a complex number $x=1+2i$. We have to prove that the expression ${{x}^{3}}+7{{x}^{2}}-13x+16$ is equal to $-29$.

In the expression ${{x}^{3}}+7{{x}^{2}}-13x+16$, we can see that ${{x}^{2}}$ and ${{x}^{3}}$ are present. So, we have to find both ${{x}^{2}}$ and ${{x}^{3}}$.

In the question, it is given $x=1+2i$.

Squaring both the sides of the above equation, we can find ${{x}^{2}}$ as,

${{x}^{2}}={{\left( 1+2i \right)}^{2}}$

We have a formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.

Using this formula to find ${{x}^{2}}$, we get,

$\begin{align}

& {{x}^{2}}={{\left( 1 \right)}^{2}}+{{\left( 2i \right)}^{2}}+2\left( 1 \right)\left( 2i \right) \\

& \Rightarrow {{x}^{2}}=1+4{{i}^{2}}+4i \\

\end{align}$

In complex numbers, we have a formula ${{i}^{2}}=-1$.

Substituting ${{i}^{2}}=-1$in the above equation, we get,

\[\begin{align}

& {{x}^{2}}=1-4+4i \\

& \Rightarrow {{x}^{2}}=-3+4i...............\left( 1 \right) \\

\end{align}\]

To find ${{x}^{3}}$, we will multiply the above equation with $x=1+2i$.

\[\begin{align}

& {{x}^{2}}.x=\left( -3+4i \right)\left( 1+2i \right) \\

& \Rightarrow {{x}^{3}}=-3-6i+4i+8{{i}^{2}} \\

& \Rightarrow {{x}^{3}}=-3-2i+8{{i}^{2}} \\

\end{align}\]

Substituting ${{i}^{2}}=-1$in the above equation, we get,

\[\begin{align}

& {{x}^{3}}=-3-2i-8 \\

& \Rightarrow {{x}^{3}}=-11-2i..........\left( 2 \right) \\

\end{align}\]

Since in the question we have to prove ${{x}^{3}}+7{{x}^{2}}-13x+16=-29$, substituting ${{x}^{2}}$

from equation $\left( 1 \right)$ and ${{x}^{3}}$ from equation $\left( 2 \right)$ in

${{x}^{3}}+7{{x}^{2}}-13x+16$, we get,

\[\begin{align}

& -11-2i+7\left( -3+4i \right)-13\left( 1+2i \right)+16 \\

& \Rightarrow -11-2i-21+28i-13-26i+16 \\

& \Rightarrow -29 \\

\end{align}\]

Hence, we have proved that ${{x}^{3}}+7{{x}^{2}}-13x+16=-29$.

Note: One can also do this question by converting the complex number $x=1+2i$ to the Euler’s form i.e. in the form of $r{{e}^{i\theta }}$ or $r\left( \cos \theta +i\sin \theta \right)$ where $r$ is the modulus of the complex number and $\theta $ is the argument of the complex number. Then using the De Moivre’s theorem i.e. ${{\left( r{{e}^{i\theta }} \right)}^{n}}=\cos n\theta +i\sin n\theta $, one can find ${{x}^{2}}$ and ${{x}^{3}}$ and substituting $x,{{x}^{2}},{{x}^{3}}$ in the expression ${{x}^{3}}+7{{x}^{2}}-13x+16$. But this method will take a lot of time since the argument of the complex number is not a standard angle. So, one has to write $\sin 2\theta ,\sin 3\theta ,\cos 2\theta ,\cos 3\theta $ in terms of $\sin \theta $ and $\cos \theta $ using the trigonometric formulas. Then, one has to simplify all the terms to get the answer.

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