Suppose z in mathbb{C} has an argument theta | vedclass

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(B) Given,$|z - 3i| = 3$,which represents a circle with radius $3$ and center $(0, 3)$.Let $z = x + iy$. Since $z$ lies on the circle $|z - 3i| = 3$,w

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$i$

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(B) Given,$|z - 3i| = 3$,which represents a circle with radius $3$ and center $(0, 3)$.Let $z = x + iy$. Since $z$ lies on the circle $|z - 3i| = 3$,we have $x^2 + (y - 3)^2 = 3^2$,which simplifies to $x^2 + y^2 - 6y = 0$.Since the argument of $z$ is $	heta$,we have $	an 	heta = \frac{y}{x}$,so $x = y \cot 	heta$.Substituting $x$ into the circle equation: $(y \cot 	heta)^2 + y^2 - 6y = 0$.$y^2(\cot^2 	heta + 1) - 6y = 0 \Rightarrow y^2 \csc^2 	heta = 6y$.Since $z 
eq 0$,$y = 6 \sin^2 	heta$.Then $x = 6 \sin^2 	heta \cdot \frac{\cos 	heta}{\sin 	heta} = 6 \sin 	heta \cos 	heta$.Thus,$z = x + iy = 6 \sin 	heta \cos 	heta + i(6 \sin^2 	heta) = 6 \sin 	heta(\cos 	heta + i \sin 	heta) = 6 \sin 	heta e^{i 	heta}$.Now,$\frac{6}{z} = \frac{6}{6 \sin 	heta e^{i 	heta}} = \frac{1}{\sin 	heta} e^{-i 	heta} = \frac{1}{\sin 	heta}(\cos 	heta - i \sin 	heta) = \cot 	heta - i$.Therefore,$\cot 	heta - \frac{6}{z} = \cot 	heta - (\cot 	heta - i) = i$.

Suppose $z \in \mathbb{C}$ has an argument $\theta$ such that $0 < \theta < \frac{\pi}{2}$ and satisfies the equation $|z - 3i| = 3$. What is the value of $\cot \theta - \frac{6}{z}$?

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