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Complex Numbers question

2019 · 10 Jan · Shift 1 · Q37
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Complex Numbers question

2019 · 10 Jan · Shift 1 · Q37

JEE MainMathematicsComplex NumbersMCQ+4 / −1
Let z1 and z2 be any two non-zero complex numbers such that 3∣z1∣=4∣z2∣.3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|.3∣z1​∣=4∣z2​∣. If z=3z12z2+2z23z1z = {{3{z_1}} \over {2{z_2}}} + {{2{z_2}} \over {3{z_1}}}z=2z2​3z1​​+3z1​2z2​​ then :
  1. A
    Im(z)=0{\rm I}m\left( z \right) = 0Im(z)=0
  2. B
    ∣z∣=172\left| z \right| = \sqrt {{17 \over 2}}∣z∣=217​​
  3. C
    ∣z∣=\left| z \right| =∣z∣= 129+16cos⁡2θ{1 \over 2}\sqrt {9 + 16{{\cos }^2}\theta }21​9+16cos2θ​
  4. D
    Re(z) === 0
View written solutionFree

Correct answer: C

  1. Given condition

We have non-zero complex numbers z1,z2z_1,z_2z1​,z2​ such that 3∣z1∣=4∣z2∣.3|z_1|=4|z_2|.3∣z1​∣=4∣z2​∣.

Also, z=3z12z2+2z23z1.z=\frac{3z_1}{2z_2}+\frac{2z_2}{3z_1}.z=2z2​3z1​​+3z1​2z2​​.

We must determine which option is correct.

  1. Write the ratio in polar form

Let w=3z12z2.w=\frac{3z_1}{2z_2}.w=2z2​3z1​​. Then z=w+1w.z=w+\frac{1}{w}.z=w+w1​.

Now compute ∣w∣|w|∣w∣ using the given condition: ∣w∣=∣3z12z2∣=3∣z1∣2∣z2∣.|w|=\left|\frac{3z_1}{2z_2}\right|=\frac{3|z_1|}{2|z_2|}.∣w∣=​2z2​3z1​​​=2∣z2​∣3∣z1​∣​. Since 3∣z1∣=4∣z2∣3|z_1|=4|z_2|3∣z1​∣=4∣z2​∣, ∣w∣=4∣z2∣2∣z2∣=2.|w|=\frac{4|z_2|}{2|z_2|}=2.∣w∣=2∣z2​∣4∣z2​∣​=2.

So we can write w=2(cos⁡θ+isin⁡θ)=2eiθw=2(\cos\theta+i\sin\theta)=2e^{i\theta}w=2(cosθ+isinθ)=2eiθ for some real θ\thetaθ.

Hence, 1w=12e−iθ=12(cos⁡θ−isin⁡θ).\frac{1}{w}=\frac{1}{2}e^{-i\theta}=\frac{1}{2}(\cos\theta-i\sin\theta).w1​=21​e−iθ=21​(cosθ−isinθ).

Therefore, z=2eiθ+12e−iθ.z=2e^{i\theta}+\frac{1}{2}e^{-i\theta}.z=2eiθ+21​e−iθ.

  1. Expand zzz into real and imaginary parts

z=2(cos⁡θ+isin⁡θ)+12(cos⁡θ−isin⁡θ).z=2(\cos\theta+i\sin\theta)+\frac{1}{2}(\cos\theta-i\sin\theta).z=2(cosθ+isinθ)+21​(cosθ−isinθ).

So, z=(2+12)cos⁡θ+i(2−12)sin⁡θ.z=\left(2+\frac12\right)\cos\theta+i\left(2-\frac12\right)\sin\theta.z=(2+21​)cosθ+i(2−21​)sinθ.

Thus, z=52cos⁡θ+32isin⁡θ.z=\frac{5}{2}\cos\theta+\frac{3}{2}i\sin\theta.z=25​cosθ+23​isinθ.

Hence,

\qquad \operatorname{Im}(z)=\frac{3}{2}\sin\theta.$$ This shows: - $\operatorname{Im}(z)$ is **not always zero**, so **A is false**. - $\operatorname{Re}(z)$ is **not always zero**, so **D is false**. --- 4. **Find $|z|$** Using the real and imaginary parts, $$|z|^2=\left(\frac{5}{2}\cos\theta\right)^2+\left(\frac{3}{2}\sin\theta\right)^2.$$ So, $$|z|^2=\frac14\left(25\cos^2\theta+9\sin^2\theta\right).$$ Using $\sin^2\theta=1-\cos^2\theta$, $$|z|^2=\frac14\left(25\cos^2\theta+9(1-\cos^2\theta)\right)$$ $$=\frac14\left(9+16\cos^2\theta\right).$$ Therefore, $$|z|=\frac12\sqrt{9+16\cos^2\theta}.$$ This matches **Option C**. --- 5. **Check option B** Option B says $$|z|=\sqrt{\frac{17}{2}}.$$ But $|z|$ depends on $\theta$, so it is not constant. Hence **B is false**. --- 6. **Conclusion** The correct option is: $$\boxed{\text{C}}$$ This agrees with the stored correct answer.
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