Question

If $|z-3i|=3$ and $argz\in (0,\frac{\pi }{2})$ then $cot\left(argz\right)-\frac{6}{z}$ is equal to ......................

• A. -i
• B. i
• C. 2i
• D. -2i

Answer 1

The correct option is B i

$|z-3i|=3$ gives
$x^2+(y-3{)}^2=9$
$x=3cosA$ and $y=3sinA+3$
Hence
$z=3[cosA+i(sinA+1\left)\right]$
$=3\left[sin\right(\frac{\pi }{2}-A)+i(1+cos(\frac{\pi }{2}-A)\left)\right]$
$=3\left[2sin\right(\frac{\pi }{4}-\frac{A}{2}\left)cos\right(\frac{\pi }{4}-\frac{A}{2}\left)\right)+i2co{s}^2(\frac{\pi }{4}-\frac{A}{2})\left)\right]$
$=6cos(\frac{\pi }{4}-\frac{A}{2})\left[sin\right(\frac{\pi }{4}-\frac{A}{2})+icos(\frac{\pi }{4}-\frac{A}{2}\left)\right]$
$=6cos(\frac{\pi }{4}-\frac{A}{2}).e^{i(\frac{\pi }{4}-\frac{A}{2})}$ .... after using $sin\theta =cos({90}^0-\theta )$
Hence
$cot\left(argz\right)$
$=cot(\frac{\pi }{4}-\frac{A}{2})$
$=tan(\frac{\pi }{4}-\frac{A}{2})$ ...(i)
and
$\frac{6}{z}$
$=sec(\frac{\pi }{4}-\frac{A}{2}).e^{-i(\frac{\pi }{4}+\frac{A}{2})}$
$=sec(\frac{\pi }{4}-\frac{A}{2})\left[sin\right(\frac{\pi }{4}-\frac{A}{2})-icos(\frac{\pi }{4}-\frac{A}{2}\left)\right]$ after using $sin\theta =cos({90}^0-\theta )$
$=tan(\frac{\pi }{4}-\frac{A}{2})-i$ ...(ii)
Hence
$i-ii$
$=-(-i)$
$=i$