If the point left(frac{k-1}{k}, frac{k-2}{k}r | vedclass

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(D) Given $\left|\frac{z+3i}{3z+i}\right| < 1$,where $z = x + iy$. Squaring both sides,we get $\frac{|z+3i|^2}{|3z+i|^2} < 1$. This implies $|x + i(y+

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$(-\infty, 1) \cup (5, \infty)$

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(D) Given $\left|\frac{z+3i}{3z+i}\right| Squaring both sides,we get $\frac{|z+3i|^2}{|3z+i|^2} This implies $|x + i(y+3)|^2 $x^2 + (y+3)^2 $x^2 + y^2 + 6y + 9 $8x^2 + 8y^2 - 8 > 0$,which simplifies to $x^2 + y^2 > 1$. Since the point $\left(\frac{k-1}{k}, \frac{k-2}{k}\right)$ satisfies this inequality: $\left(\frac{k-1}{k}\right)^2 + \left(\frac{k-2}{k}\right)^2 > 1$. $\frac{k^2 - 2k + 1 + k^2 - 4k + 4}{k^2} > 1$. $2k^2 - 6k + 5 > k^2$. $k^2 - 6k + 5 > 0$. $(k-1)(k-5) > 0$. Thus,$k \in (-\infty, 1) \cup (5, \infty)$.

If the point $\left(\frac{k-1}{k}, \frac{k-2}{k}\right)$ lies on the locus of $z$ satisfying the inequality $\left|\frac{z+3i}{3z+i}\right| < 1$,then the interval in which $k$ lies is

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