# What are solutions to equation z^2= conj z?

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### 1 Answer

You need to consider the complex number `z = a + b*i` , hence, its complex conjugate `bar z = a - b*i` .

You need to solve the given equation `z^2 = bar z` , replacing `a + b*i` for `z` and `a - b*i` for `bar z` , such that:

`(a + b*i)^2 = (a - b*i)`

Squaring to the left side, yields:

`a^2 + 2a*b*i + b^2*i^2 = a - b*i`

Since `i^2 = -1` yields:

`a^2 + 2ab*i - b^2 = a - b*i`

Equating the real and imaginary parts both sides, yields:

`{(a^2 - b^2 = a),(2ab = -b):} => {(a^2 - b^2 = a),(2a = -1):} => {(a^2 - b^2 = a),(a = -1/2):} `

Replacing -`1/2` for a in the top equation yields:

`(-1/2)^2 - b^2 = -1/2 => 1/4 - b^2 = -1/2 => -b^2 = -1/4 - 1/2`

`b^2 = (1/4 + 1/2) => b^2 = 3/4 => b_(1,2) = +-sqrt(3/4)`

`b_(1,2) = +-sqrt3/2`

**Hence, evaluating the solutions to the given equation z^2 = bar z yields `z_1 = -1/2 + sqrt3/2*i` and **`z_2 = -1/2 - sqrt3/2*i.`