Question

A particle starts to travel from a pt. 'P' on curve $C_1:|z-3-4i|=5$, where $|z|\to$ max. From 'P', the particle moves through an $\angle$ of $ta{n}^{-1}\left(\frac{3}{4}\right)$ in Anti-CW direction on $C_2$ and reaches Q. From Q, it comes down parallel to $I_m$-axis by 2 units & reaches R.
Find the co-ordinates of R.

Answer 1

Above expression basically re[present circle whose center $(3,4)$ & redii $5$ from figure $C\left(z_1\right)=3+4i$

$\tan \alpha =\frac{4}{3}$$

$]\beta ={\tan }^{-1}\left(\frac{3}{4}\right)$

$\tan \beta =\frac{3}{4}$

$\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha +\tan \beta }$

$\tan (\alpha +\beta )=\frac{\frac{4}{3}+\frac{3}{4}}{1-\frac{4}{3}.\frac{3}{4}}=\frac{\frac{4}{3}+\frac{3}{4}}{0}=\infty$

$\tan (\alpha +\beta )=\tan \pi /2$

$\alpha +\beta =\pi /2$

co-ordinary of $Q=(3,CM+r)$

$=(3,4+5)$

$=(3,9)$

we known

$\sin \alpha =\frac{CM}{OC}$

$\frac{4}{5}=\frac{CM}{5}$

$CM=4$

It start $2$ until down

Co-ordinant of $R=(3+7i)$ or (3,7)