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In triangle ABC with vertices A(z1), B(z2), C(z3), the angle ABC = pi/4 and AB/BC = sqrt(2). Express z2 in terms of z1 and z3.

1. z2 = (z1 + z3)/2
2. z2 = z1 + i(z3 - z1)
3. z2 = (1/(1 - i))*z1 + (-i/(1 - i))*z3
4. z2 = z1 + (z3 - z1)/sqrt(2)

Correct answer: z2 = (1/(1 - i))*z1 + (-i/(1 - i))*z3

Solution

Rotating BC to BA: (z1 - z2)/(z3 - z2) = (AB/BC)e^(i pi/4) = sqrt(2)*(1/sqrt2)(1 + i) = 1 + i; solving gives z2 = (z1 - i z3)/(1 - i).

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