Find the two square roots for the following complex number. Write your answers in standard form. (Enter your answers as a comma-separated list.)−2 + 2i3

To find the square roots of the complex number $-2 + 2i$, we'll follow the required steps:

1. Break Down the Problem

We want to find $z$ such that $z^2 = -2 + 2i$. We can express $z$ in standard form as $z = x + yi$, where $x$ and $y$ are real numbers.

2. Relevant Concepts

Using the formula for squaring a complex number, $(x + yi)^2 = x^2 - y^2 + 2xyi$ We need to equate this to the complex number $-2 + 2i$. This gives us two equations to solve:

1. $x^2 - y^2 = -2$
2. $2xy = 2$

3. Analysis and Detail

From the second equation: $2xy = 2 \implies xy = 1 \implies y = \frac{1}{x}$

Substituting $y$ into the first equation: $x^2 - \left(\frac{1}{x}\right)^2 = -2$ Multiply through by $x^2$ to eliminate the fraction: $x^4 + 2x^2 - 1 = 0$ Let $u = x^2$, we then have: $u^2 + 2u - 1 = 0$ Using the quadratic formula $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $u = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}$ This gives us two potential values for $u$:

1. $u_1 = -1 + \sqrt{2}$
2. $u_2 = -1 - \sqrt{2}$ (not possible since $u$ must be non-negative)

Thus, $u = -1 + \sqrt{2}$, which is positive.

Now substituting back to find $x$: $x^2 = -1 + \sqrt{2} \implies x = \pm \sqrt{-1 + \sqrt{2}}$

Now substituting for $y$: $y = \frac{1}{x} = \pm \frac{1}{\sqrt{-1 + \sqrt{2}}}$

4. Verify and Summarize

Thus, the solutions for $z = x + yi$ are:

1. $z_1 = \sqrt{-1 + \sqrt{2}} + \frac{i}{\sqrt{-1 + \sqrt{2}}}$
2. $z_2 = -\sqrt{-1 + \sqrt{2}} - \frac{i}{\sqrt{-1 + \sqrt{2}}}$

Final Answer

The two square roots of the complex number $-2 + 2i$ in standard form are: $\sqrt{-1 + \sqrt{2}} + \frac{i}{\sqrt{-1 + \sqrt{2}}}, -\sqrt{-1 + \sqrt{2}} - \frac{i}{\sqrt{-1 + \sqrt{2}}}$