𝐐𝐮𝐞𝐬𝐭𝐢𝐨𝐧: Find the square root of complex number -8-6i

𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:

Let the square root of -8 - 6i be x + yi, where x and y are real numbers.

Then, (x + yi)² = -8 - 6i

Expanding the left side, we get:

x² + 2xyi + (yi)² = -8 - 6i

x² + 2xyi - y² = -8 - 6i

(x² - y²) + (2xy)i = -8 - 6i

Equating the real and imaginary parts, we get two equations:

1) x² - y² = -8

2) 2xy = -6 = xy = -3

From the second equation, y = -3/x. Substitute this into the first equation:

x² - (-3/x)² = -8

x² - 9/x² = -8

Multiply the entire equation by x²:

x⁴ - 9 = -8x²

x⁴ + 8x² - 9 = 0

This is a quadratic equation in x². Let u = x². Then the equation becomes:

u² + 8u - 9 = 0

Factor the quadratic:

(u + 9)(u - 1) = 0

So, u = -9 or u = 1.

Since x is a real number, x² must be non-negative. Therefore, x² = 1, which gives x = ±1.

Now find the corresponding values of y using y = -3/x:

If x = 1, y = -3/1 = -3.

If x = -1, y = -3/(-1) = 3.

Thus, the square roots of -8 - 6i are 1 - 3i and -1 + 3i. We can write this as ±(1 - 3i).