Chapter 9.5, Problem 69E

To determine

To calculate: Expansion of ${(2-3i)}^6$ by Binomial Theorem.

Answer to Problem 69E

The Binomial expansion of ${(2-3i)}^6$ is given by $2035+828i$ .

Explanation of Solution

Given: ${(2-3i)}^6$

Formula used:The Binomial Theorem is given by,

  ${(a+b)}^n=C{}_0{a}^n{b}^0+C{}_1{a}^{n-1}{b}^1+C{}_2{a}^{n-2}{b}^2+\mathrm{...........}+C{}_n{a}^0{b}^n$

where $C{}_r=\frac{n!}{(n-r)!r!}$

Calculation:

  ${(2-3i)}^6$

  $=C{}_0{2}^6{(-3i)}^0+C{}_1{2}^{6-1}{(-3i)}^1+C{}_2{2}^{6-2}{(-3i)}^2+C{}_3{2}^{6-3}{(-3i)}^3+C{}_4{2}^{6-4}{(-3i)}^4+C{}_5{2}^{6-5}{(-3i)}^5+C{}_6{2}^0{(-3i)}^6$

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

  $\begin{array}{l}=\frac{1}{1}\cdot 2^6\cdot 1+\frac{6}{1}\cdot 2^5\cdot (-3i)+\frac{15}{1}\cdot 2^4\cdot (-3^2)+\frac{20}{1}\cdot 2^3\cdot (3^{3}i)+\frac{15}{1}\cdot 2^2\cdot 3^4+\frac{6}{1}\cdot 2^1\cdot (-3^{5}i)+\frac{1}{1}\cdot 2^0\cdot (-3^6)[\because i^2=-1]\\ =64-576i-2160+4320i+4860-2916i-729\\ =(64-2160+4860-729)+(-576+4320-2916)i\\ =2035+828i\end{array}$

Conclusion:

The Binomial expansion of ${(2-3i)}^6$ is given by $2035+828i$ .