Tip:A–D to answerE for explanationV for videoS to reveal answer
Let z be a complex number such that z+1z−2i=2,z=i. Then z lies on the circle of radius 2 and centre
- A.
(2,0)
- B.
(0,2)
- C.
(0,−2)
(Correct Answer) - D.
(0,0)
Explanation
Given:
Let z=p+iq
⇒p2+(q−2)2=(2)2[p2+(q+1)2]
This is the equation of the circle of the form x2+y2+2gx+2fy+c=0, where centre≡(−g,−f) and radius=a2=g2+f2−C
So, here g=0,f=2
Explanation
Given:
Let z=p+iq
⇒p2+(q−2)2=(2)2[p2+(q+1)2]
This is the equation of the circle of the form x2+y2+2gx+2fy+c=0, where centre≡(−g,−f) and radius=a2=g2+f2−C
So, here g=0,f=2
