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Let $\theta = \arg{z}$ and suppose $z$ satisfies $|z - 3i| = 3$.

Compute all possible values of $\displaystyle \cot{\theta} - \frac{6}{z}$. Note that $\displaystyle 0 < \theta < \frac{\pi}{2}$.

$\bf{My\; Try::}$ Let $z-3i=3e^{i\theta}\Rightarrow z=3i+3\cos \theta+3i\sin \theta = 3\cos \theta+3i(1+\sin \theta)$

So we get $$z=3\sin \left(\frac{\pi}{2}-\theta\right)+3i\left[1+\cos\left(\frac{\pi}{2}-\theta\right)\right]$$

So we get $$z=3i\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)e^{-i\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}$$

Now how can i solve after that, Thanks

juantheron
  • 53,015

2 Answers2

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Since $\theta=\arg z$, note that $z=3\cos\theta+3i(1+\sin\theta)$ is not correct.

Let $z=x+yi$ where $x,y\in\mathbb R$. Then, since we have $$x^2+(y-3)^2=3^2$$we can write $$x=3\cos\alpha,\qquad y=3+3\sin\alpha$$ where $-\pi/2\lt\alpha\lt \pi/2$. So, we can have $$\cos\theta=\frac{x}{\sqrt{x^2+y^2}}=\frac{3\cos\alpha}{\sqrt{6(3+3\sin\alpha)}}$$ $$\sin\theta=\frac{y}{\sqrt{x^2+y^2}}=\frac{3+3\sin\alpha}{\sqrt{6(3+3\sin\alpha)}}$$ and so $$\begin{align}\cot\theta-\frac{6}{z}&=\frac{3\cos\alpha}{3+3\sin\alpha}-\frac{6}{3\cos\alpha+(3+3\sin\alpha)i}\\&=\frac{\cos\alpha}{1+\sin\alpha}-\frac{2(\cos\alpha-(1+\sin\alpha)i)}{\cos^2\alpha+(1+\sin\alpha)^2}\\&=\frac{\cos\alpha}{1+\sin\alpha}-\frac{\cos\alpha-(1+\sin\alpha)i}{1+\sin\alpha}\\&=\frac{1+\sin\alpha}{1+\sin\alpha}i\\&=\color{red}{i}\end{align}$$

mathlove
  • 139,939
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WLOG $z-3i=3(\cos2t+i\sin2t)$

$\iff z=3\{\cos2t+i(1+\sin2t)\}$ $=6\sin\left(\dfrac\pi4-t\right)\left[\cos\left(\dfrac\pi4-t\right)+i\sin\left(\dfrac\pi4-t\right)\right]$

$\dfrac6z=\cdots=\dfrac{\cos\left(\dfrac\pi4-t\right)-i\sin\left(\dfrac\pi4-t\right)}{\sin\left(\dfrac\pi4-t\right)}=\cot\left(\dfrac\pi4-t\right)-i$

$\tan\theta=\cdots=\tan\left(\dfrac\pi4-t\right)$

Can you take it from here?

lab bhattacharjee
  • 274,582