Few special problems on square root of a complex number

Finding the Square Root of a Complex Expression

Given the complex number:

z = 3a + i√(a⁴ - 7a² + 1)

We aim to find √z. Let:

√z = x + iy

Then, squaring both sides:

z = (x + iy)² = x² - y² + 2ixy

Comparing real and imaginary parts:

x² - y² = 3a        ...(1)
2xy = √(a⁴ - 7a² + 1)  ...(2)
  

From (2):

xy = ½√(a⁴ - 7a² + 1)

Let S = x² + y² and D = x² - y² = 3a. Then:

x² = (S + D)/2
y² = (S - D)/2
x²y² = [(S² - D²)/4] = ¼(a⁴ - 7a² + 1)
  

Substitute D = 3a:

S² - 9a² = a⁴ - 7a² + 1
⇒ S² = a⁴ + 2a² + 1 = (a² + 1)²
⇒ S = a² + 1
  

Now compute x² and y²:

x² = (a² + 1 + 3a)/2 = (a² + 3a + 1)/2
y² = (a² + 1 - 3a)/2 = (a² - 3a + 1)/2
  

So the square root of z is:

√z = √[(a² + 3a + 1)/2] + i√[(a² - 3a + 1)/2]
  

Note: The sign may vary based on quadrant conventions.