# How do you use the binomial theorem to expand (1+i)^4?

Jun 7, 2017

${\left(1 + i\right)}^{4} = - 4$

#### Explanation:

We know ${\left(a + b\right)}^{n} = n {C}_{0} {a}^{n} \cdot {b}^{0} + n {C}_{1} {a}^{n - 1} \cdot {b}^{1} + n {C}_{2} {a}^{n - 2} \cdot {b}^{2} + \ldots \ldots \ldots . + n {C}_{n} {a}^{n - n} \cdot {b}^{n}$

Here $a = 1 , b = i , n = 4$ We know, nC_r = (n!)/(r!*(n-r)!
$\therefore 4 {C}_{0} = 1 , 4 {C}_{1} = 4 , 4 {C}_{2} = 6 , 4 {C}_{3} = 4 , 4 {C}_{4} = 1$
${i}^{2} = - 1 , {i}^{3} = - i , {i}^{4} = 1$

$\therefore {\left(1 + i\right)}^{4} = {1}^{4} + 4 \cdot {1}^{3} \cdot i + 6 \cdot {1}^{2} \cdot {i}^{2} + 4 \cdot 1 \cdot {i}^{3} + {i}^{4}$ or

${\left(1 + i\right)}^{4} = 1 + 4 \cdot i + 6 \cdot {i}^{2} + 4 \cdot {i}^{3} + {i}^{4}$ or

${\left(1 + i\right)}^{4} = 1 + \cancel{4 i} - 6 - \cancel{4 i} + 1 = - 4$ [Ans]