Chapter 8.4, Problem 89E

To determine

To expand: the complex number.

Answer to Problem 89E

The complex number is $161+252i$ .

Explanation of Solution

Given information:

Given expression

  ${\left(i+4\right)}^4$

Calculation:

The binomial theorem is stated below

In the expansion of ${\left(x+y\right)}^n$

  ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\mathrm{...}{+}_n{C}_r{x}^{n-r}{y}^r+\mathrm{...}+nx{y}^{n-1}+y^n$

The coefficient of $x^{n-r}{y}^r$ is

  ${}_n{C}_r=\frac{n!}{r!\left(n-r\right)!}$

The symbol $\left(\begin{array}{l}n\\ r\end{array}\right)$ is often used in place of ${}_n{C}_r$ to denote binomial coefficients.

According to the binomial theorem

In the expansion of ${\left(x+y\right)}^n$

  ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\mathrm{...}{+}_n{C}_r{x}^{n-r}{y}^r+\mathrm{...}+nx{y}^{n-1}+y^n$

If we substitute $i$ for x, 4 for $y$ , and 4 for $n$ , we get ${\left(i+4\right)}^4$

From the above expansion

  $\begin{array}{c}{\left(i+4\right)}^4{=}_4{C}_0{\left(i\right)}^4{\left(4\right)}^0{+}_4{C}_1{\left(i\right)}^3\left(4\right){+}_4{C}_2{\left(4\right)}^2{i}^2{+}_4{C}_3{\left(4\right)}^{3}i+{\left(4\right)}^4\\ =1+4\left(-i\right)+\frac{4\times 3}{2}\left(-1\right)\times 16+4\times 64\left(i\right)+256\\ =1-4i-96+256i+256\\ =161+252i\end{array}$