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Complex Numbers question

2023 · 30 Jan · Shift 1 · Q35
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  5. /2023 · 30 Jan · Shift 1 · Q35

Complex Numbers question

2023 · 30 Jan · Shift 1 · Q35

JEE MainMathematicsComplex NumbersNumerical+4 / −1
Let z=1+iz=1+iz=1+i and z1=1+izˉzˉ(1−z)+1zz_{1}=\frac{1+i \bar{z}}{\bar{z}(1-z)+\frac{1}{z}}z1​=zˉ(1−z)+z1​1+izˉ​. Then 12πarg⁡(z1)\frac{12}{\pi} \arg \left(z_{1}\right)π12​arg(z1​) is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 9

  1. Given values

We have z=1+iz=1+iz=1+i and hence zˉ=1−i.\bar z=1-i.zˉ=1−i.

We need to compute z1=1+izˉzˉ(1−z)+1zz_1=\frac{1+i\bar z}{\bar z(1-z)+\frac1z}z1​=zˉ(1−z)+z1​1+izˉ​ then find 12πarg⁡(z1).\frac{12}{\pi}\arg(z_1).π12​arg(z1​).

  1. Compute the numerator

First, izˉ=i(1−i)=i−i2=i+1=1+i.i\bar z=i(1-i)=i-i^2=i+1=1+i.izˉ=i(1−i)=i−i2=i+1=1+i. So, 1+izˉ=1+(1+i)=2+i.1+i\bar z=1+(1+i)=2+i.1+izˉ=1+(1+i)=2+i.

Thus numerator =2+i=2+i=2+i.

  1. Compute the denominator

(a) Find 1−z1-z1−z

1−z=1−(1+i)=−i.1-z=1-(1+i)=-i.1−z=1−(1+i)=−i.

So, zˉ(1−z)=(1−i)(−i)=−i+i2=−i−1=−1−i.\bar z(1-z)=(1-i)(-i)=-i+i^2=-i-1=-1-i.zˉ(1−z)=(1−i)(−i)=−i+i2=−i−1=−1−i.

(b) Find 1z\dfrac1zz1​

1z=11+i=1−i(1+i)(1−i)=1−i2.\frac1z=\frac{1}{1+i}=\frac{1-i}{(1+i)(1-i)}=\frac{1-i}{2}.z1​=1+i1​=(1+i)(1−i)1−i​=21−i​.

(c) Add them

Therefore denominator is (−1−i)+1−i2.(-1-i)+\frac{1-i}{2}.(−1−i)+21−i​. Writing over common denominator 222, −2−2i+1−i2=−1−3i2.\frac{-2-2i+1-i}{2}=\frac{-1-3i}{2}.2−2−2i+1−i​=2−1−3i​.

So denominator =−1−3i2=\dfrac{-1-3i}{2}=2−1−3i​.

  1. Compute z1z_1z1​

z1=2+i(−1−3i)/2=2(2+i)−1−3i=4+2i−1−3i.z_1=\frac{2+i}{(-1-3i)/2}=\frac{2(2+i)}{-1-3i}=\frac{4+2i}{-1-3i}.z1​=(−1−3i)/22+i​=−1−3i2(2+i)​=−1−3i4+2i​.

Now rationalize: z1=(4+2i)(−1+3i)(−1−3i)(−1+3i).z_1=\frac{(4+2i)(-1+3i)}{(-1-3i)(-1+3i)}.z1​=(−1−3i)(−1+3i)(4+2i)(−1+3i)​.

Denominator: (−1)2+32=1+9=10.(-1)^2+3^2=1+9=10.(−1)2+32=1+9=10.

Numerator:

(4+2i)(−1+3i)=−4+12i−2i+6i2=−4+10i−6=−10+10i.(4+2i)(-1+3i)= -4+12i-2i+6i^2=-4+10i-6=-10+10i.(4+2i)(−1+3i)=−4+12i−2i+6i2=−4+10i−6=−10+10i.

Hence z1=−10+10i10=−1+i.z_1=\frac{-10+10i}{10}=-1+i.z1​=10−10+10i​=−1+i.

  1. Find the argument of z1z_1z1​

Since z1=−1+i,z_1=-1+i,z1​=−1+i, it lies in the second quadrant, and its principal argument is arg⁡(z1)=3π4.\arg(z_1)=\frac{3\pi}{4}.arg(z1​)=43π​.

  1. Evaluate the required expression

12πarg⁡(z1)=12π⋅3π4=9.\frac{12}{\pi}\arg(z_1)=\frac{12}{\pi}\cdot\frac{3\pi}{4}=9.π12​arg(z1​)=π12​⋅43π​=9.

  1. Comparison with stored answer

Our derived answer is 999, which matches the stored correct answer.

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