Question

Find the value of (1 + i)⁴ in the form of a +ib:​

Answer 1

$\huge\mathfrak\orange{Answer}$

$(1 + {i})^{4} \\ = ((1 - { {i)}^{2} )}^{2} \\ = {(1 - i)}^{2} ( {1 - i)}^{2} \\ = ( {1}^{2} + {i}^{2} - 2 \times 1 \times i) \\ ( {1}^{2} + {i}^{2} - 2 \times 1 \times i) \\ \\ = (1 + {i}^{2} - 2i)(1 + {i}^{2} - 2i) \\ put \: {i}^{2} = - 1 \\ = (1 - 1 - 2i)(1 - 1 - 2i) \\ = (0 - 2i)(0 - 2i) \\ = ( - 2i)( - 2i) \\ = 4 {i}^{2} \\ putting \: {i}^{2} = - 1 \\ = 4( - 1) \\ = - 4 \\ = - 4 + 0 \\ = - 4 + 0i \\ a = - 4 \\ b = 0$

$a + ib = - 4 + 0i$