Question

if 1+4√3i = (a+ib)^2
prove that a^2 - b^2 = 1 and ab = 2√3​

Answer 1

$1 + 4 \sqrt{3} i = ( {a + ib})^{2}...(1) \\ taking \: conjugate \: of \: euation \: \: (1) \\ \\ 1 - 4 \sqrt{3} i = ( {a - ib})^{2} ...(2) \\ \\ (1) + (2) \to \\ \\ ( {a + ib})^{2} + ( {a - ib})^{2} = 2 \\ \implies \: {a}^{2} - {b}^{2} + 2iab + {a}^{2} - {b}^{2} - 2iab = 2 \\ \implies \: 2( {a}^{2} - {b}^{2} ) = 2 \\ \implies \: {a}^{2} - {b}^{2} = 1 \\ \\ \\ (1) - (2) \to \\ \\ ( {a + ib})^{2} - ( {a - ib})^{2} = 1 + 4 \sqrt{3} i - 1 + 4 \sqrt{3} i \\ \implies \: {a}^{2} - {b}^{2} + 2iab - {a}^{2} + {b}^{2} + 2iab = 8i \sqrt{3} \\ \implies \: 4iab = 8 \sqrt{3} i \\ \implies \: ab = 2 \sqrt{3}$