The square roots of the complex number z=3+4i are:

We are given the complex number z = 3 + 4i and asked to find its square roots , i.e., complex numbers w = x + iy such that w² = z . Let the square root be w = x + iy We want: (x + iy)² = 3 + 4i Expanding the left-hand side: x² + 2ixy + i²y² = x² – y² + 2ixy (because i² = –1) So, Real part: x² – y² = 3 Imaginary part: 2xy = 4 Solve the system From 2xy = 4 → xy = 2 From x² – y² = 3 Now use xy = 2 to express y in terms of x: y = 2/x Substitute into x² – y² = 3: x² – (2/x)² = 3 x² – 4/x² = 3 Multiply both sides by x² to eliminate the denominator: x⁴ – 4 = 3x² Rearranging: x⁴ – 3x² – 4 = 0 Let u = x², then: u² – 3u – 4 = 0 Solve using quadratic formula: u = [3 ± √(9 + 16)] / 2 = [3 ± √25]/2 = [3 ± 5]/2 So, u = 4 or u = –1 But u = x² must be non-negative , so discard –1. Therefore, x² = 4 → x = ±2 Now, recall y = 2/x:

• If x = 2, y = 1 → w = 2 + i
• If x = –2, y = –1 → w = –2 – i

The square roots of 3 + 4i are: ± (2 + i)