Finding the Square Root of i

Step 1: Represent the problem

We want z such that:

z² = i

where z ∈ ℂ.

Write z = x + iy with x, y ∈ ℝ.

Step 2: Expand and equate

(x + iy)² = x² − y² + 2xyi

We want this equal to i = 0 + 1·i.

x² − y² + 2xy i = 0 + 1·i

Equating real and imaginary parts:

Real part: x² − y² = 0 ⇒ x² = y²

Imaginary part: 2xy = 1

Step 3: Solve the system

From x² = y², we have y = ±x.

Case 1: y = x

From 2xy = 1:

2x² = 1 ⇒ x² = ½ ⇒ x = ±1/√2

Then y = x, so:

z = (1/√2) + i(1/√2) or z = −(1/√2) − i(1/√2)

Case 2: y = −x

From 2xy = 1:

2x(−x) = 1 ⇒ −2x² = 1 ⇒ x² = −½

No real solution for x, so this case is invalid.

Step 4: Simplify and verify

From x = 1/√2, y = 1/√2, we get:

z = (1 + i)/√2 = e^{iπ/4}

The other root is:

z = −(1 + i)/√2 = −e^{iπ/4}

Check:

(e^{iπ/4})² = e^{iπ/2} = i

(−e^{iπ/4})² = e^{iπ/2} = i

Both are valid square roots of i.

The two square roots of i are:

(1 + i)/√2 and −(1 + i)/√2

Equivalently: e^iπ/4 and e^i5π/4.