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