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Complex Numbers question

2019 · 10 Apr · Shift 1 · Q27
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Complex Numbers question

2019 · 10 Apr · Shift 1 · Q27

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If a > 0 and z = (1+i)2a−i{{{{\left( {1 + i} \right)}^2}} \over {a - i}}a−i(1+i)2​, has magnitude 25\sqrt {{2 \over 5}}52​​, then z‾\overline zz is equal to :
  1. A
    −15+35i- {1 \over 5} + {3 \over 5}i−51​+53​i
  2. B
    −15−35i- {1 \over 5} - {3 \over 5}i−51​−53​i
  3. C
    15−35i{1 \over 5} - {3 \over 5}i51​−53​i
  4. D
    −35−15i- {3 \over 5} - {1 \over 5}i−53​−51​i
View written solutionFree

Correct answer: B

  1. Simplify the given complex number

We have

z=(1+i)2a−i,a>0z = \frac{(1+i)^2}{a-i}, \qquad a>0z=a−i(1+i)2​,a>0

Now,

(1+i)2=1+2i+i2=1+2i−1=2i(1+i)^2 = 1+2i+i^2 = 1+2i-1 = 2i(1+i)2=1+2i+i2=1+2i−1=2i

So,

z=2ia−iz = \frac{2i}{a-i}z=a−i2i​
  1. Use the given magnitude

Given that

∣z∣=25|z|=\sqrt{\frac{2}{5}}∣z∣=52​​

But

∣z∣=∣2i∣∣a−i∣=2a2+1|z|=\frac{|2i|}{|a-i|} = \frac{2}{\sqrt{a^2+1}}∣z∣=∣a−i∣∣2i∣​=a2+1​2​

Therefore,

2a2+1=25\frac{2}{\sqrt{a^2+1}}=\sqrt{\frac{2}{5}}a2+1​2​=52​​

Squaring both sides:

4a2+1=25\frac{4}{a^2+1}=\frac{2}{5}a2+14​=52​ 20=2(a2+1)20=2(a^2+1)20=2(a2+1) 10=a2+110=a^2+110=a2+1 a2=9a^2=9a2=9

Since a>0a>0a>0, we get

a=3a=3a=3
  1. Find zzz

Substitute a=3a=3a=3:

z=2i3−iz=\frac{2i}{3-i}z=3−i2i​

Rationalize the denominator:

z=2i(3+i)(3−i)(3+i)z=\frac{2i(3+i)}{(3-i)(3+i)}z=(3−i)(3+i)2i(3+i)​ z=6i+2i29+1z=\frac{6i+2i^2}{9+1}z=9+16i+2i2​ z=6i−210z=\frac{6i-2}{10}z=106i−2​ z=−15+35iz=-\frac{1}{5}+\frac{3}{5}iz=−51​+53​i
  1. Find the conjugate

Therefore,

z‾=−15−35i\overline z=-\frac{1}{5}-\frac{3}{5}iz=−51​−53​i
  1. Match with the options

This corresponds to:

Option B:

−15−35i-\frac{1}{5}-\frac{3}{5}i−51​−53​i
  1. Compare with stored correct answer

Stored correct answer: B

Our derived answer is also B, so they agree.

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