The point z lies on the circle |z−2i|=22 then arg⁡z−2z+2 is equal to

x2+(y−2)2=8x2+y2−4y=4arg⁡z−2z+2=tan−1⁡yx−2−tan−1⁡yx+2=tan−1⁡yx−2−yx+21+yx−2⋅yx+2=tan−1⁡4yx2+y2−4=tan−1⁡4y4y=tan−1⁡1=π4⇒ (c)

The point z lies on the circle |z−2i|=22 then arg⁡z−2z+2 is equal to

x2+(y−2)2=8x2+y2−4y=4arg⁡z−2z+2=tan−1⁡yx−2−tan−1⁡yx+2=tan−1⁡yx−2−yx+21+yx−2⋅yx+2=tan−1⁡4yx2+y2−4=tan−1⁡4y4y=tan−1⁡1=π4⇒ (c)

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

The point $z$ lies on the circle $| z - 2 i | = 2 \sqrt{2}$ then $\arg (\frac{z - 2}{z + 2})$ is equal to

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

$\frac{π}{3}$

$\frac{π}{4}$

$\frac{π}{6}$

$\frac{π}{8}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{aligned} x^{2} + (y - 2)^{2} = 8 \\ x^{2} + y^{2} - 4 y = 4 \end{aligned}$

$\arg (\frac{z - 2}{z + 2}) = \tan^{- 1} \frac{y}{x - 2} - \tan^{- 1} \frac{y}{x + 2}$

$\begin{array}{l} = \tan^{- 1} \frac{\frac{y}{x - 2} - \frac{y}{x + 2}}{1 + \frac{y}{x - 2} \cdot \frac{y}{x + 2}} \\ = \tan^{- 1} \frac{4 y}{x^{2} + y^{2} - 4} \\ = \tan^{- 1} \frac{4 y}{4 y} = \tan^{- 1} 1 = \frac{π}{4} \Rightarrow (c) \end{array}$