Question

Solve the equation `z^2 = barz`, where z = x + iy.

Answer 1

`z^2 = barz`

⇒ x² – y² + i2xy = x – iy

Therefore, x² – y² = x  ......(1)

And 2xy = –y  ......(2)

From (2), we have y = 0 or x = `- 1/2`

When y = 0, from (1)

We get x² – x = 0

i.e., x = 0 or x = 1.

When x = `-1/2`, from (1)

We get y² = `1/4 + 1/2` or y² = `3/4`.

i.e., y² = `+- sqrt(3)/2`

Hence, the solutions of the given equation are 0 + i0, 1 + i0, `-1/2 +i sqrt(3)/2`, `-1/2 -i sqrt(3)/2`