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Complex Numbers question

2019 · 11 Jan · Shift 2 · Q34
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Complex Numbers question

2019 · 11 Jan · Shift 2 · Q34

JEE MainMathematicsComplex NumbersMCQ+4 / −1
Let z be a complex number such that |z| + z = 3 + i (where i = −1\sqrt { - 1}−1​). Then |z| is equal to :
  1. A
    343{{\sqrt {34} } \over 3}334​​
  2. B
    53{5 \over 3}35​
  3. C
    54{5 \over 4}45​
  4. D
    414{{\sqrt {41} } \over 4}441​​
View written solutionFree

Correct answer: B

  1. Let z=x+iyz=x+iyz=x+iy where x,y∈Rx,y\in \mathbb{R}x,y∈R.

  2. Then ∣z∣=x2+y2|z|=\sqrt{x^2+y^2}∣z∣=x2+y2​ and the given condition is ∣z∣+z=3+i.|z|+z=3+i.∣z∣+z=3+i.

  3. Substitute z=x+iyz=x+iyz=x+iy: x2+y2+x+iy=3+i.\sqrt{x^2+y^2}+x+iy=3+i.x2+y2​+x+iy=3+i.

  4. Compare real and imaginary parts:

    • Imaginary part: y=1y=1y=1
    • Real part: x2+1+x=3\sqrt{x^2+1}+x=3x2+1​+x=3

  5. Let r=∣z∣=x2+1.r=|z|=\sqrt{x^2+1}.r=∣z∣=x2+1​. Then the real-part equation becomes r+x=3⇒x=3−r.r+x=3 \quad \Rightarrow \quad x=3-r.r+x=3⇒x=3−r.

  6. Also, since r=∣z∣r=|z|r=∣z∣, r2=x2+1.r^2=x^2+1.r2=x2+1. Substitute x=3−rx=3-rx=3−r: r2=(3−r)2+1.r^2=(3-r)^2+1.r2=(3−r)2+1.

  7. Expand: r2=9−6r+r2+1r^2=9-6r+r^2+1r2=9−6r+r2+1 r2=r2+10−6rr^2=r^2+10-6rr2=r2+10−6r 0=10−6r0=10-6r0=10−6r 6r=106r=106r=10 r=53.r=\frac{5}{3}.r=35​.

  8. Therefore, ∣z∣=53.|z|=\frac{5}{3}.∣z∣=35​.

  9. Check with options:

    • A: 343\dfrac{\sqrt{34}}{3}334​​
    • B: 53\dfrac{5}{3}35​
    • C: 54\dfrac{5}{4}45​
    • D: 414\dfrac{\sqrt{41}}{4}441​​

    So the correct option is B.

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