Let s, t, r be non-zero complex numbers and L | vedclass

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(A) Let $z = x + iy$,$s = s_1 + is_2$,$t = t_1 + it_2$,and $r = r_1 + ir_2$.The equation $sz + t\bar{z} + r = 0$ becomes:$(s_1 + is_2)(x + iy) + (t_1

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$A, B, C, D$

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(A) Let $z = x + iy$,$s = s_1 + is_2$,$t = t_1 + it_2$,and $r = r_1 + ir_2$.The equation $sz + t\bar{z} + r = 0$ becomes:$(s_1 + is_2)(x + iy) + (t_1 + it_2)(x - iy) + (r_1 + ir_2) = 0$$(s_1x - s_2y + t_1x + t_2y + r_1) + i(s_2x + s_1y - t_2x + t_1y + r_2) = 0$Equating real and imaginary parts to zero:$(s_1 + t_1)x + (t_2 - s_2)y + r_1 = 0$$(s_2 - t_2)x + (s_1 + t_1)y + r_2 = 0$This is a system of two linear equations in $x$ and $y$. The determinant of the coefficient matrix is $D = (s_1 + t_1)^2 - (t_2 - s_2)(s_2 - t_2) = (s_1 + t_1)^2 + (t_2 - s_2)^2 = |s + t|^2$.If $|s| 
eq |t|$,then $D 
eq 0$ (unless $s = -t$),leading to a unique solution (a point).If $|s| = |t|$,the lines are either parallel or coincident. Specifically,if $s 
eq -t$,the system represents a line or is inconsistent.Thus,if $L$ has more than one element,it must be a line (infinitely many elements).$L$ is either a point or a line. The intersection of a line and a circle is at most $2$ points. If $L$ is a point,the intersection is at most $1$ point. Thus,$C$ is true.Therefore,statements $A, B, C, D$ are all true.

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