Question

Calcular raiz quadrada de 3-4i

Answer 1

Seja a + bi a raiz

$\sqrt{3-4i} =a+bi$

Quadrando, 3 - 4i = (a + bi)² => 3-4i = a² + 2abi +b²i²

a² + 2abi +b²(-1) = 3 - 4i => a² - b² + 2abi = 3  - 4i

$\left \{ {{a^2-b^2=3} \atop {2ab=-4}} \right. \\ \\ ab=-2=\ \textgreater \ b=-2/a \\ \\ a^2-(-2/a)^2=-2=\ \textgreater \ a^2-4/a^2=3=\ \textgreater \ a ^{4}-3a^2-4=0$

Δ = (-3)²-4.1.(-4)

Δ = 9 + 16 = 25

a² = (3-5)/2 = -1 (não serve) ou

a² = ( 3 + 5)/2 = 4 =>a² = 4 => a = -2 ou a = 2

p/ a = -2 => b = -2/(-2) = 1

p/ a = 2 => b = -2/2 = -1

√(3-4i) = -2 + i ou √(3-4i) = 2 - i