Chapter 9.5, Problem 73E

To determine

To expand: the complex number.

Answer to Problem 73E

The complex number is $-4$ .

Explanation of Solution

Given information:

Given equation

  ${\left(1+i\right)}^4$

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 $y$ and 4 for $n$ , we get ${\left(1+i\right)}^4$

From the above expansion

  $\begin{array}{c}{\left(1+i\right)}^4=1^4{+}_4{C}_1{1}^{3}i{+}_4{C}_2{1}^2{i}^2{+}_4{C}_{3}1i^3+i^4\\ =1+4i+\frac{4\times 3}{2}\left(-1\right)+4\left(-i\right)+1\\ =-4\end{array}$