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Complex Numbers question

2019 · 10 Jan · Shift 2 · Q28
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Complex Numbers question

2019 · 10 Jan · Shift 2 · Q28

JEE MainMathematicsComplex NumbersMCQ+4 / −1
Let z=(32+i2)5+(32−i2)5.z = {\left( {{{\sqrt 3 } \over 2} + {i \over 2}} \right)^5} + {\left( {{{\sqrt 3 } \over 2} - {i \over 2}} \right)^5}.z=(23​​+2i​)5+(23​​−2i​)5. If R(z) and 1(z) respectively denote the real and imaginary parts of z, then :
  1. A
    R(z) = −-− 3
  2. B
    R(z) < 0 and I(z) > 0
  3. C
    I(z) = 0
  4. D
    R(z) > 0 and I(z) > 0
View written solutionFree

Correct answer: C

  1. Write each complex number in trigonometric form

Given

z=(32+i2)5+(32−i2)5.z=\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^5+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^5.z=(23​​+2i​)5+(23​​−2i​)5.

Notice that

32+i2=cos⁡π6+isin⁡π6=cis⁡π6,\frac{\sqrt{3}}{2}+\frac{i}{2}=\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}=\operatorname{cis}\frac{\pi}{6},23​​+2i​=cos6π​+isin6π​=cis6π​,

and

32−i2=cos⁡π6−isin⁡π6=cos⁡(−π6)+isin⁡(−π6)=cis⁡(−π6).\frac{\sqrt{3}}{2}-\frac{i}{2}=\cos\frac{\pi}{6}-i\sin\frac{\pi}{6}=\cos\left(-\frac{\pi}{6}\right)+i\sin\left(-\frac{\pi}{6}\right)=\operatorname{cis}\left(-\frac{\pi}{6}\right).23​​−2i​=cos6π​−isin6π​=cos(−6π​)+isin(−6π​)=cis(−6π​).

So,

z=(cis⁡π6)5+(cis⁡(−π6))5.z=\left(\operatorname{cis}\frac{\pi}{6}\right)^5+\left(\operatorname{cis}\left(-\frac{\pi}{6}\right)\right)^5.z=(cis6π​)5+(cis(−6π​))5.
  1. Apply De Moivre's theorem

Using

(cis⁡θ)n=cis⁡(nθ),(\operatorname{cis}\theta)^n=\operatorname{cis}(n\theta),(cisθ)n=cis(nθ),

we get

z=cis⁡5π6+cis⁡(−5π6).z=\operatorname{cis}\frac{5\pi}{6}+\operatorname{cis}\left(-\frac{5\pi}{6}\right).z=cis65π​+cis(−65π​).

That is,

z=(cos⁡5π6+isin⁡5π6)+(cos⁡5π6−isin⁡5π6).z=\left(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right)+\left(\cos\frac{5\pi}{6}-i\sin\frac{5\pi}{6}\right).z=(cos65π​+isin65π​)+(cos65π​−isin65π​).
  1. Add the two terms

The imaginary parts cancel:

z=2cos⁡5π6.z=2\cos\frac{5\pi}{6}.z=2cos65π​.

Now,

cos⁡5π6=−32.\cos\frac{5\pi}{6}=-\frac{\sqrt{3}}{2}.cos65π​=−23​​.

Hence,

z=2(−32)=−3.z=2\left(-\frac{\sqrt{3}}{2}\right)=-\sqrt{3}.z=2(−23​​)=−3​.
  1. Find real and imaginary parts

Since

z=−3,z=-\sqrt{3},z=−3​,

we have

R(z)=−3,I(z)=0.R(z)=-\sqrt{3}, \qquad I(z)=0.R(z)=−3​,I(z)=0.
  1. Check the options

  • A: R(z)=−3R(z)=-3R(z)=−3
    False, because R(z)=−3R(z)=-\sqrt{3}R(z)=−3​.

  • B: R(z)>0R(z)>0R(z)>0 (the printed option appears as R(z)  0R(z)\; 0R(z)0, which in this standard question is clearly intended as R(z)>0R(z)>0R(z)>0)
    False, since −3<0-\sqrt{3}<0−3​<0.

  • C: I(z)=0I(z)=0I(z)=0
    True.

  • D: R(z)>0R(z)>0R(z)>0 and I(z)>0I(z)>0I(z)>0
    False.

Therefore, the correct option is C.

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