If z is a complex number such that z-iz-1 is purely imaginary, then the minimum value of z - 3 + 3i is

z-iz-1=x+iy-ix+iy-1=x+i(y-1)(x-1)+iy z-iz-1 =(x+i(y-1))((x-1)-iy)((x-1)+iy)((x-1)-iy) z-iz-1=x(x-1)+y(y-1)(x-1)2+y2+((x-1)(y-1)-xy)(x-1)2+y2is purely imaginary⇒ Real part = 0 x(x-1)+y(y-1)=0 ⇒ x2 + y2 -x - y = 0z lies on a circle with centre c12, 12 and r = 14 + 14 = 12min of z - 3 + 3i = min distance from P(3, 3) from the circle=CP-r =3-122+3-122-12 =254+254-12=252-12 =52-12=42=22

If z is a complex number such that z-iz-1 is purely imaginary, then the minimum value of z - 3 + 3i is

z-iz-1=x+iy-ix+iy-1=x+i(y-1)(x-1)+iy z-iz-1 =(x+i(y-1))((x-1)-iy)((x-1)+iy)((x-1)-iy) z-iz-1=x(x-1)+y(y-1)(x-1)2+y2+((x-1)(y-1)-xy)(x-1)2+y2is purely imaginary⇒ Real part = 0 x(x-1)+y(y-1)=0 ⇒ x2 + y2 -x - y = 0z lies on a circle with centre c12, 12 and r = 14 + 14 = 12min of z - 3 + 3i = min distance from P(3, 3) from the circle=CP-r =3-122+3-122-12 =254+254-12=252-12 =52-12=42=22

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If z is a complex number such that $\frac{z - i}{z - 1}$ is purely imaginary, then the minimum value of $|z - (3 + 3 i)|$ is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$2 \sqrt{2}$

$6 \sqrt{2}$

$2 \sqrt{2} - 1$

$3 \sqrt{2}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\frac{z - i}{z - 1} = \frac{x + iy - i}{x + iy - 1} = \frac{x + i (y - 1)}{(x - 1) + iy} \frac{z - i}{z - 1} = \frac{(x + i (y - 1)) ((x - 1) - iy)}{((x - 1) + iy) ((x - 1) - iy)} \frac{z - i}{z - 1} = \frac{x (x - 1) + y (y - 1)}{(x - 1)^{2} + y^{2}} + \frac{((x - 1) (y - 1) - xy)}{(x - 1)^{2} + y^{2}}$

is purely imaginary

$\Rightarrow Real part = 0$ $x (x - 1) + y (y - 1) = 0 \Rightarrow x^{2} + y^{2} - x - y = 0$

z lies on a circle with centre $c (\frac{1}{2}, \frac{1}{2})$ and $r = \sqrt{\frac{1}{4} + \frac{1}{4}} = \frac{1}{\sqrt{2}}$

min of $|z - (3 + 3 i)| =$ min distance from P(3, 3) from the circle

$= C P - r = \sqrt{{(3 - \frac{1}{2})}^{2} + {(3 - \frac{1}{2})}^{2}} - \frac{1}{\sqrt{2}} = \sqrt{\frac{25}{4} + \frac{25}{4}} - \frac{1}{\sqrt{2}} = \sqrt{\frac{25}{2}} - \frac{1}{\sqrt{2}} = \frac{5}{\sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2 \sqrt{2}$