11. Find the square roots of the complex number \( 2+2 \sqrt{3} i \), leaving your answer in rectangular form. (Hint: Solve \( z^{2}=2+2 \sqrt{3} i \), where \( z=a+b i \) and \( a \) and \( b \) are real numbers.)

To find the square roots of the complex number \( 2 + 2\sqrt{3}i \), we assume \( z = a + bi \), where \( a \) and \( b \) are real numbers. Squaring \( z \) gives: \[ z^2 = (a + bi)^2 = a^2 + 2abi - b^2 = (a^2 - b^2) + (2ab)i \] We want to match the real and imaginary parts with \( 2 + 2\sqrt{3}i \): 1. Real part: \( a^2 - b^2 = 2 \) 2. Imaginary part: \( 2ab = 2\sqrt{3} \) From the second equation, we can simplify it to find \( ab = \sqrt{3} \). Now we have a system of equations: 1. \( a^2 - b^2 = 2 \) 2. \( ab = \sqrt{3} \) Using \( b = \frac{\sqrt{3}}{a} \) (from \( ab = \sqrt{3} \)), substitute \( b \) in the first equation: \[ a^2 - \left(\frac{\sqrt{3}}{a}\right)^2 = 2 \] This simplifies to: \[ a^2 - \frac{3}{a^2} = 2 \] Multiplying through by \( a^2 \) to eliminate the fraction: \[ a^4 - 2a^2 - 3 = 0 \] Now, let \( x = a^2 \). We transform the equation into: \[ x^2 - 2x - 3 = 0 \] Factoring, we get: \[ (x - 3)(x + 1) = 0 \] This gives \( x = 3 \) or \( x = -1 \) (we discard \( x = -1 \) as \( a^2 \) cannot be negative). Thus, \( a^2 = 3 \) which leads to \( a = \sqrt{3} \) or \( a = -\sqrt{3} \). Using \( ab = \sqrt{3} \), if \( a = \sqrt{3} \): \[ b = \frac{\sqrt{3}}{\sqrt{3}} = 1 \] If \( a = -\sqrt{3} \): \[ b = \frac{\sqrt{3}}{-\sqrt{3}} = -1 \] Thus, the square roots of \( 2 + 2\sqrt{3}i \) in rectangular form are: \[ \sqrt{3} + i \quad \text{and} \quad -\sqrt{3} - i. \]