Question

$z_1,z_2,z_3,z_4$ are the complex numbers satisfying $|z-13i|=r.$ $z_1,z_2$ correspond to maximum and minimum moduli, whereas $z_3,z_4$ represent maximum and minimum arguments. If $z_1-z_2=24i,$ then $z_3-z_4$ is equal to

• A. $\frac{120}{13}$
• B. $-\frac{120}{13}$
• C. $10$
• D. $-1$

Answer 1

The correct option is B $-\frac{120}{13}$

$|z-13i|=r$ represents a circle in argand plane centred at $(0,13)$, and radius '$r$'
$z_1,z_2$ corresponds to max. & min. modulii lying on this circle.
$\therefore z_1-z_2=24i$ (given)
$\therefore (13+r)i-(13-r)i=24i$
$\therefore 2ri=24i$
$\therefore r=12$ - radius of circle.
In $\Delta OAB,B=$ origin,
$O=$centre of circle
$A=$point of contact of tangent from 'B'
by pythagoras theorem
$AB=5$
$A$ is the point on circle : ${}^{\prime }{z}_4^{\prime }$ having minimum argument.
Reflection of 'A' cony 'y'-axis i.e. ${}^{\prime }{z}_3^{\prime }$ has maximum argument.

$\therefore$ Coordinates of $A(=z_4)=5(\frac{12}{13}+\frac{i5}{13})$

$\therefore$ Coordinates of $A^{\prime }(-z_3)=5(\frac{-12}{13}+\frac{i5}{13})$

$\therefore z_3-z_4=(\frac{-60}{13}+\frac{25}{13})-(\frac{60}{13}+\frac{25i}{13})=\frac{-120}{13}$