#One complex solution

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random oarBOT
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frosty island
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Hm, interesting. Let's see.
Suppose z = a + bi and k = u + vi. Then we have:
a^2 - b^2 + 2abi + a^2 + b^2 + a - bi + u + vi - 2i = 0
Simplifying a bit, we get:
(2a^2 + a + u) + (2ab - b + v - 2)i = 0
So, we get a system:
2a^2 + a + u = 0
2ab - b + v - 2 = 0
The discriminant of the first equation is D = 1 - 8k, so there will be only one solution when k = 1/8. Let's check whether that solution exists.
2a^2 + a + 1/8 = 0
a = -1/4
From the second equation:
(2a - 1)b = -(v - 2)
b = -(v + 2)/(2a - 1) = -(v - 2)/(-3/2) = (2/3)(v - 2)
So, there is a unique solution z = -1/4 + (2/3)(v - 2)i for each k = 1/8 + vi.

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Though, I guess they wanted k to be real. In that case only k = 1/8 works.