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Complex Numbers question

2021 · 16 Mar · Shift 1 · Q42
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Complex Numbers question

2021 · 16 Mar · Shift 1 · Q42

JEE MainMathematicsComplex NumbersNumerical+4 / −1
Let z and ω\omegaω be two complex numbers such that ω=zz‾−2z+2,∣z+iz−3i∣=1\omega = z\overline z - 2z + 2,\left| {{{z + i} \over {z - 3i}}} \right| = 1ω=zz−2z+2,​z−3iz+i​​=1 and Re(ω\omegaω) has minimum value. Then, the minimum value of n ∈\in∈ N for which ω\omegaω n is real, is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 4

  1. Interpret the condition

Given

ω=zzˉ−2z+2\omega = z\bar z - 2z + 2ω=zzˉ−2z+2

and

∣z+iz−3i∣=1.\left|\frac{z+i}{z-3i}\right|=1.​z−3iz+i​​=1.

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

Then

∣z+iz−3i∣=1  ⟹  ∣z+i∣=∣z−3i∣.\left|\frac{z+i}{z-3i}\right|=1 \implies |z+i|=|z-3i|.​z−3iz+i​​=1⟹∣z+i∣=∣z−3i∣.

So the point zzz is equidistant from −i-i−i and 3i3i3i in the Argand plane.

  1. Find the locus of zzz

Now, z+i=x+i(y+1),z−3i=x+i(y−3).z+i=x+i(y+1), \qquad z-3i=x+i(y-3).z+i=x+i(y+1),z−3i=x+i(y−3). Hence

∣z+i∣2=x2+(y+1)2,|z+i|^2=x^2+(y+1)^2,∣z+i∣2=x2+(y+1)2, ∣z−3i∣2=x2+(y−3)2.|z-3i|^2=x^2+(y-3)^2.∣z−3i∣2=x2+(y−3)2.

Equating,

x2+(y+1)2=x2+(y−3)2.x^2+(y+1)^2=x^2+(y-3)^2.x2+(y+1)2=x2+(y−3)2.

So,

(y+1)2=(y−3)2.(y+1)^2=(y-3)^2.(y+1)2=(y−3)2.

Expanding,

y2+2y+1=y2−6y+9y^2+2y+1=y^2-6y+9y2+2y+1=y2−6y+9 8y=88y=88y=8 y=1.y=1.y=1.

Thus the locus is the horizontal line Im⁡(z)=1,\operatorname{Im}(z)=1,Im(z)=1, so z=x+i.z=x+i.z=x+i.

  1. Compute ω\omegaω in terms of xxx

Since z=x+iz=x+iz=x+i,

∣z∣2=zzˉ=x2+1.|z|^2=z\bar z=x^2+1.∣z∣2=zzˉ=x2+1.

Therefore

ω=zzˉ−2z+2=(x2+1)−2(x+i)+2.\omega=z\bar z-2z+2=(x^2+1)-2(x+i)+2.ω=zzˉ−2z+2=(x2+1)−2(x+i)+2.

Simplify:

ω=x2−2x+3−2i.\omega=x^2-2x+3-2i.ω=x2−2x+3−2i.

Hence,

Re⁡(ω)=x2−2x+3=(x−1)2+2.\operatorname{Re}(\omega)=x^2-2x+3=(x-1)^2+2.Re(ω)=x2−2x+3=(x−1)2+2.
  1. Minimize Re⁡(ω)\operatorname{Re}(\omega)Re(ω)

Clearly,

Re⁡(ω)=(x−1)2+2\operatorname{Re}(\omega)=(x-1)^2+2Re(ω)=(x−1)2+2

is minimum when x=1.x=1.x=1.

So the corresponding value of zzz is z=1+i,z=1+i,z=1+i, and then

ω=12−2(1)+3−2i=2−2i.\omega=1^2-2(1)+3-2i=2-2i.ω=12−2(1)+3−2i=2−2i.
  1. Find minimum n∈Nn\in\mathbb Nn∈N such that ωn\omega^nωn is real

We have

ω=2−2i=22(cos⁡(−π4)+isin⁡(−π4)).\omega=2-2i=2\sqrt2\left(\cos\left(-\frac\pi4\right)+i\sin\left(-\frac\pi4\right)\right).ω=2−2i=22​(cos(−4π​)+isin(−4π​)).

Then

ωn=(22)n(cos⁡(−nπ4)+isin⁡(−nπ4)).\omega^n=(2\sqrt2)^n\left(\cos\left(-\frac{n\pi}{4}\right)+i\sin\left(-\frac{n\pi}{4}\right)\right).ωn=(22​)n(cos(−4nπ​)+isin(−4nπ​)).

For ωn\omega^nωn to be real, its argument must be a multiple of π\piπ:

−nπ4=kπ-\frac{n\pi}{4}=k\pi−4nπ​=kπ

for some integer kkk. Thus

n4=−k⇒n=4(−k).\frac n4=-k \quad\Rightarrow\quad n=4(-k).4n​=−k⇒n=4(−k).

So nnn must be a multiple of 444.

The smallest positive natural number is

4.\boxed{4}.4​.
  1. Comparison with stored answer

Derived answer: 444. Stored correct answer: 444. They agree.

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