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Complex Numbers question

2017 · 9 Apr · Shift 1 · Q28
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Complex Numbers question

2017 · 9 Apr · Shift 1 · Q28

JEE MainMathematicsComplex NumbersMCQ+4 / −1
The equation Im (iz−2z−i)\left( {{{iz - 2} \over {z - i}}} \right)(z−iiz−2​)+ 1 = 0, z ∈\in∈ C, z eee i represents a part of a circle having radius equal to :
  1. A
    2
  2. B
    1
  3. C
    34{3 \over 4}43​
  4. D
    12{1 \over 2}21​
View written solutionFree

Correct answer: C

  1. Let z=x+iy,  x,y∈R,z≠i.z=x+iy,\,\, x,y\in\mathbb R, \quad z\ne i.z=x+iy,x,y∈R,z=i.

  2. Compute the expression iz−2z−i.\frac{iz-2}{z-i}.z−iiz−2​. First, iz=i(x+iy)=ix−y,iz=i(x+iy)=ix-y,iz=i(x+iy)=ix−y, so iz−2=(−y−2)+ix.iz-2=(-y-2)+ix.iz−2=(−y−2)+ix. Also, z−i=x+i(y−1).z-i=x+i(y-1).z−i=x+i(y−1).

  3. Rationalize the denominator: iz−2z−i=((−y−2)+ix)(x−i(y−1))x2+(y−1)2.\frac{iz-2}{z-i}=\frac{((-y-2)+ix)(x-i(y-1))}{x^2+(y-1)^2}.z−iiz−2​=x2+(y−1)2((−y−2)+ix)(x−i(y−1))​.

  4. Expand the numerator carefully. Let A=−y−2,B=x,C=x,D=−(y−1).A=-y-2,\quad B=x,\quad C=x,\quad D=-(y-1).A=−y−2,B=x,C=x,D=−(y−1). Then (A+iB)(C+iD)=(AC−BD)+i(AD+BC).(A+iB)(C+iD)=(AC-BD)+i(AD+BC).(A+iB)(C+iD)=(AC−BD)+i(AD+BC).

    So the imaginary part of the numerator is AD+BC=(−y−2)(−(y−1))+x⋅x.AD+BC=(-y-2)(-(y-1))+x\cdot x.AD+BC=(−y−2)(−(y−1))+x⋅x. Simplify: (−y−2)(−(y−1))=(y+2)(y−1)=y2+y−2.(-y-2)(-(y-1))=(y+2)(y-1)=y^2+y-2.(−y−2)(−(y−1))=(y+2)(y−1)=y2+y−2. Hence ℑ(numerator)=x2+y2+y−2.\Im\text{(numerator)}=x^2+y^2+y-2.ℑ(numerator)=x2+y2+y−2.

    Therefore, ℑ(iz−2z−i)=x2+y2+y−2x2+(y−1)2.\Im\left(\frac{iz-2}{z-i}\right)=\frac{x^2+y^2+y-2}{x^2+(y-1)^2}.ℑ(z−iiz−2​)=x2+(y−1)2x2+y2+y−2​.

  5. Use the given equation: ℑ(iz−2z−i)+1=0\Im\left(\frac{iz-2}{z-i}\right)+1=0ℑ(z−iiz−2​)+1=0 ⇒ℑ(iz−2z−i)=−1.\Rightarrow \Im\left(\frac{iz-2}{z-i}\right)=-1.⇒ℑ(z−iiz−2​)=−1.

    Thus, x2+y2+y−2x2+(y−1)2=−1.\frac{x^2+y^2+y-2}{x^2+(y-1)^2}=-1.x2+(y−1)2x2+y2+y−2​=−1.

  6. Cross-multiply: x2+y2+y−2=−(x2+(y−1)2).x^2+y^2+y-2=-(x^2+(y-1)^2).x2+y2+y−2=−(x2+(y−1)2). Expand the right side: x2+y2+y−2=−x2−(y2−2y+1).x^2+y^2+y-2=-x^2-(y^2-2y+1).x2+y2+y−2=−x2−(y2−2y+1). x2+y2+y−2=−x2−y2+2y−1.x^2+y^2+y-2=-x^2-y^2+2y-1.x2+y2+y−2=−x2−y2+2y−1.

    Bring all terms to one side: 2x2+2y2−y−1=0.2x^2+2y^2-y-1=0.2x2+2y2−y−1=0.

  7. Divide by 2: x2+y2−y2−12=0.x^2+y^2-\frac y2-\frac12=0.x2+y2−2y​−21​=0.

  8. Complete the square in yyy: x2+(y2−y2)−12=0x^2+\left(y^2-\frac y2\right)-\frac12=0x2+(y2−2y​)−21​=0 x2+(y−14)2−116−12=0x^2+\left(y-\frac14\right)^2-\frac1{16}-\frac12=0x2+(y−41​)2−161​−21​=0 x2+(y−14)2=116+816=916.x^2+\left(y-\frac14\right)^2=\frac1{16}+\frac8{16}=\frac9{16}.x2+(y−41​)2=161​+168​=169​.

  9. This is a circle with center (0,14)\left(0,\frac14\right)(0,41​) and radius r=34.r=\frac34.r=43​.

  10. Since z≠iz\ne iz=i, the point i=(0,1)i=(0,1)i=(0,1) is excluded. Indeed, 02+(1−14)2=(34)2,0^2+\left(1-\frac14\right)^2=\left(\frac34\right)^2,02+(1−41​)2=(43​)2, so it lies on the circle. Hence the locus is a part of this circle, but the radius remains 34.\boxed{\frac34}.43​​.

Therefore, the correct option is C.

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