To find the square root of $i$, we solve for $z$ such that:

$z^2 = i$

Step 1: Represent $z$ in polar form

Assume $z = a + bi$, where $a$ and $b$ are real numbers. Then:

$z^2 = (a + bi)^2 = a^2 - b^2 + 2abi$

For $z^2 = i$, we match real and imaginary parts:

1. $a^2 - b^2 = 0$ (real part)
2. $2ab = 1$ (imaginary part)

Step 2: Solve the equations

From $a^2 - b^2 = 0$, we get $a^2 = b^2$, so $a = \pm b$.

Case 1: $a = b$

If $a = b$, substitute into $2ab = 1$:

$2a^2 = 1 \implies a^2 = \frac{1}{2} \implies a = \pm \frac{\sqrt{2}}{2}$

Case 2: $a = -b$

If $a = -b$, substitute into $2ab = 1$:

$2a(-a) = 1 \implies -2a^2 = 1 \implies a^2 = -\frac{1}{2}$

This case is invalid since $a^2$ cannot be negative for real $a$.

Step 3: Polar form solution

Alternatively, write $i$ in polar form:

$i = e^{i\pi/2}$

The square root of a complex number $r e^{i\theta}$ is:

$\sqrt{r} e^{i\theta/2}$

For $i = e^{i\pi/2}$, the square roots are:

$\sqrt{i} = e^{i\pi/4} = \cos\frac{\pi}{4} + i\sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}$

and

$\sqrt{i} = e^{i(5\pi/4)} = \cos\frac{5\pi}{4} + i\sin\frac{5\pi/4} = -\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}$

Final Answer:

$\sqrt{i} = \pm \left(\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}\right)$

Let me know if you want further details or have any questions!

Expansion Questions:

1. What is the geometric interpretation of the square root of $i$?
2. How can the polar form simplify complex number calculations?
3. What are other roots (e.g., cube roots) of $i$?
4. How is the modulus of a complex number involved in finding roots?
5. Can this method apply to other non-real complex numbers?

Tip:

Always convert complex numbers to polar