Question

1+4√3i find the square root of complex numbers​

Answer 1

Answer:

± (2 + √3i)

Step-by-step explanation:

Let 1 + 4√3i = (a + bi)² → √(1 + 4√3i) = a + bi

=> 1 + 4√3i = a² + b²i² + 2abi

=> 1 + 4√3i = a² - b² + 2abi

Real number RHS = real number LHS

=> 1 = a² - b²

Same for imaginary numbers,

=> 4√3 = 2ab → 2√3 =ab → 2√3/a = b

Hence,

=> 1 = a² - b² → 1 = a² - (2√3/a)²

=> a² = a⁴ - 12 → a⁴ - a² - 12 = 0

=> a⁴ - 4a² + 3a² - 12 = 0

=> a²(a² - 4) + 3(a² - 4) = 0

=> (a² - 4)(a² + 3) = 0

=> a = ± 2 or a = √3i, it is better to ignore a = √3i, after all we will get same result.

So, b = ± 2√3/2 = ± √3

Hence,

a + bi = √(1 + 4√3i)

2 + √3i or - 2 - √3i = √(1 + 4√3i)

2 + √3i or - (2 + √3i) = √(1 + 4√3i)

± (2 + √3i) = √(1 + 4√3i)