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Complex Numbers question

2020 · 9 Jan · Shift 2 · Q25
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Complex Numbers question

2020 · 9 Jan · Shift 2 · Q25

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If z be a complex number satisfying |Re(z)| + |Im(z)| = 4, then |z| cannot be :
  1. A
    10\sqrt {10}10​
  2. B
    7\sqrt {7}7​
  3. C
    172\sqrt {{{17} \over 2}}217​​
  4. D
    8\sqrt {8}8​
View written solutionFree

Correct answer: B

  1. Let z=x+iyz=x+iyz=x+iy where x=Re⁡(z)x=\operatorname{Re}(z)x=Re(z) and y=Im⁡(z)y=\operatorname{Im}(z)y=Im(z).

Given, ∣x∣+∣y∣=4|x|+|y|=4∣x∣+∣y∣=4 We need to find possible values of ∣z∣=x2+y2|z|=\sqrt{x^2+y^2}∣z∣=x2+y2​ So let r=∣z∣=x2+y2r=|z|=\sqrt{x^2+y^2}r=∣z∣=x2+y2​ with constraint ∣x∣+∣y∣=4|x|+|y|=4∣x∣+∣y∣=4.

  1. Put a=∣x∣,b=∣y∣a=|x|,\quad b=|y|a=∣x∣,b=∣y∣ Then a,b≥0,a+b=4a,b\ge 0, \quad a+b=4a,b≥0,a+b=4 and r2=x2+y2=a2+b2r^2=x^2+y^2=a^2+b^2r2=x2+y2=a2+b2 So the problem reduces to finding possible values of a2+b2a^2+b^2a2+b2 when a+b=4a+b=4a+b=4.

  2. Using (a+b)2=a2+b2+2ab(a+b)^2=a^2+b^2+2ab(a+b)2=a2+b2+2ab we get 16=a2+b2+2ab16=a^2+b^2+2ab16=a2+b2+2ab Hence a2+b2=16−2aba^2+b^2=16-2aba2+b2=16−2ab Now since a,b≥0a,b\ge 0a,b≥0 and a+b=4a+b=4a+b=4, we have 0≤ab≤40\le ab\le 40≤ab≤4 Maximum of ababab occurs at a=b=2a=b=2a=b=2, giving ab=4ab=4ab=4. Minimum of ababab occurs at (a,b)=(4,0)(a,b)=(4,0)(a,b)=(4,0) or (0,4)(0,4)(0,4), giving ab=0ab=0ab=0.

Therefore, a2+b2∈[16−2⋅4, 16−0]=[8,16]a^2+b^2 \in [16-2\cdot 4,\,16-0]=[8,16]a2+b2∈[16−2⋅4,16−0]=[8,16] So, r2∈[8,16]r^2 \in [8,16]r2∈[8,16] and hence r∈[8,4]r \in [\sqrt{8},4]r∈[8​,4]

  1. Now check the options:

  • A: 10\sqrt{10}10​ 10∈[8,16]10\in [8,16]10∈[8,16] possible.

  • B: 7\sqrt{7}7​ 7∉[8,16]7\notin [8,16]7∈/[8,16] not possible.

  • C: 172\sqrt{\frac{17}{2}}217​​ 172=8.5∈[8,16]\frac{17}{2}=8.5\in [8,16]217​=8.5∈[8,16] possible.

  • D: 8\sqrt{8}8​ this is the minimum value, attained when a=b=2a=b=2a=b=2, i.e. ∣x∣=∣y∣=2|x|=|y|=2∣x∣=∣y∣=2. So possible.

  1. Hence the value that ∣z∣|z|∣z∣ cannot be is 7\boxed{\sqrt{7}}7​​ which corresponds to option B.
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