Chapter 8.4, Problem 93E

To determine

To expand: the complex number.

Answer to Problem 93E

The complex number is $2035+828i$ .

Explanation of Solution

Given information:

Given equation

  ${\left(2-3i\right)}^6$

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 $2$ for x, $-3i$ for $y$ , and 6 for $n$ , we get ${\left(2-3i\right)}^6$

From the above expansion

  ${(2-3i)}^6=C{}_0{2}^6{(-3i)}^0+C{}_1{2}^{6-1}{(-3i)}^1+C{}_2{2}^{6-2}{(-3i)}^2+C{}_3{2}^{6-3}{(-3i)}^3$

  $+C{}_4{2}^{6-4}{(-3i)}^4+C{}_5{2}^{6-5}{(-3i)}^5+C{}_6{2}^0{(-3i)}^6$

  $=\frac{6!}{6!0!}{2}^6{(-3i)}^0+\frac{6!}{5!1!}{2}^5{(-3i)}^1+\frac{6!}{4!2!}{2}^4{(-3i)}^2+\frac{6!}{3!3!}{2}^3{(-3i)}^3$

  $+\frac{6!}{2!4!}{2}^2{(-3i)}^4+\frac{6!}{1!5!}{2}^1{(-3i)}^5+\frac{6!}{0!6!}{2}^0{(-3i)}^6$

  $=\frac{1}{1}\cdot 2^6\cdot 1+\frac{6}{1}\cdot 2^5\cdot (-3i)+\frac{15}{1}\cdot 2^4\cdot (-3^2)+\frac{20}{1}\cdot 2^3\cdot (3^{3}i)+\frac{15}{1}\cdot 2^2\cdot 3^4+\frac{6}{1}\cdot 2^1\cdot (-3^{5}i)+\frac{1}{1}\cdot 2^0\cdot (-3^6)[\because i^2=-1]$

  $=64-576i-2160+4320i+4860-2916i-729$

  $=(64-2160+4860-729)+(-576+4320-2916)i$

  $=2035+828i$