Let omega = z bar{z} + k_1 z + k_2 i z + lamb | vedclass

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(A) Given $\omega = z \bar{z} + k_1 z + k_2 i z + \lambda(1 + i)$. Let $z = x + iy$.Then $\omega = (x^2 + y^2) + k_1(x + iy) + k_2 i(x + iy) + \lambda

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$110$

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(A) Given $\omega = z \bar{z} + k_1 z + k_2 i z + \lambda(1 + i)$. Let $z = x + iy$.Then $\omega = (x^2 + y^2) + k_1(x + iy) + k_2 i(x + iy) + \lambda + i\lambda = (x^2 + y^2 + k_1 x - k_2 y + \lambda) + i(k_1 y + k_2 x + \lambda)$.$\operatorname{Re}(\omega) = x^2 + y^2 + k_1 x - k_2 y + \lambda = 0$.The circle $C$ has radius $1$,touches $y = 1$ and the $y$-axis $(x = 0)$ in the first quadrant,so its center is $(1, 1)$.Comparing $x^2 + y^2 + k_1 x - k_2 y + \lambda = 0$ with $(x - 1)^2 + (y - 1)^2 = 1^2$,we get $x^2 + y^2 - 2x - 2y + 1 = 0$.Thus $k_1 = -2, k_2 = 2, \lambda = 1$.$\operatorname{Im}(\omega) = k_1 y + k_2 x + \lambda = -2y + 2x + 1 = 0$,or $2x - 2y + 1 = 0$.The distance $d$ from the center $(1, 1)$ to the line $2x - 2y + 1 = 0$ is $d = \frac{|2(1) - 2(1) + 1|}{\sqrt{2^2 + (-2)^2}} = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}}$.The length of the chord $AB$ is $2\sqrt{r^2 - d^2} = 2\sqrt{1^2 - \frac{1}{8}} = 2\sqrt{\frac{7}{8}} = 2\frac{\sqrt{7}}{2\sqrt{2}} = \sqrt{\frac{7}{2}}$.$(AB)^2 = \frac{7}{2} = 3.5$.$30(AB)^2 = 30 	imes 3.5 = 105$.

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