Question

Square root of 1 +iota

Answer 1

Answer:

$let \: \sqrt{(1 + i)} = a + ib \\ square \: both \: side \\ 1 + i = {a}^{2} + {i}^{2} {b}^{2} + 2aib \\ 1 + i = {a}^{2} - {b}^{2} + 2aib \\ then \\ {a}^{2} - {b}^{2} = 1 \\ 2ab = 1 \\ w e\: know \: that \\ {( {a}^{2} + {b}^{2} )}^{2} = {( {a}^{2} - {b}^{2} )}^{2} + 4 {a}^{4} {b}^{4} \\ = {(1)}^{2} + 4 \times {1}^{4} \\ = 1 + 4 \\ = 5 \\ then \: {a}^{2} + {b}^{2} = \sqrt{5} \\ {a}^{2} - {b}^{2} = 1 \\ 2 {a}^{2} = \sqrt{5} + 1 \\ {a}^{2} = \frac{ \sqrt{5} + 1 }{2} \\ a = \sqrt{ \frac{ \sqrt{5} + 1}{2} }$

now put the values and find the value of

a+ib