Home
Class 12
MATHS
If complex number z lies on the curve |z...

If complex number z lies on the curve `|z - (- 1+ i)| = 1`, then find the locus of the complex number `w =(z+i)/(1-i), i =sqrt-1`.

Text Solution

Verified by Experts

The correct Answer is:
A circle with centre at `(-3//2,1//2)` and radius `1//sqrt(2)`

We have`|z + 1 -i|= 1" "(1)`
and `w = (z +i)/(1-i)`
`therefore w ( 1- i) = z +i`
` rArr z = w (1-i) -i`
Putting the value of z in (1), we get
`|w (1-i) -i + 1 -i|=1`
`rArr | 1-i||w +(1-2i)/(1-i)|=1`
`rArr |w+(3-i)/(2)| = (1)/(sqrt(2))`
`rArr |w + (-3 +i)/(2)| = (1)/(sqrt(2))`
Therefore, locus of w is a c ircle havig centre at `(-3//2),1//2)` and radius `(1)/(sqrt(2))`.
Promotional Banner

Topper's Solved these Questions

  • COMPLEX NUMBERS

    CENGAGE|Exercise Exercise 3.10|10 Videos
  • COMPLEX NUMBERS

    CENGAGE|Exercise Exercise 3.11|6 Videos
  • COMPLEX NUMBERS

    CENGAGE|Exercise Exercise 3.8|11 Videos
  • CIRCLES

    CENGAGE|Exercise Question Bank|32 Videos
  • CONIC SECTIONS

    CENGAGE|Exercise Solved Examples And Exercises|91 Videos

Similar Questions

Explore conceptually related problems

If |z-1-i|=1 , then the locus of a point represented by the complex number 5(z-i)-6 is

Find the complex number z for which |z| = z + 1 + 2i.

If a complex number z satisfies z + sqrt(2) |z + 1| + i=0 , then |z| is equal to :

The locus of a complex number z satisfying |z-(1+3i)|+|z+3-6i|=4

The complex number z satisfying the equation |z-i|=|z+1|=1

Find the complex number z for which |z+1| = z + 2(1 + i) .

find the difference of the complex numbers: z_(1)=-3+2i and z_(2)=13-i .

The conjugate of a complex number z is (2)/(1-i) . Then Re(z) equals

If omega = z//[z-(1//3)i] and |omega| = 1 , then find the locus of z.

If z is a complex number satisfying the relation |z+1|=z+2(1+i), then z is