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Complex Numbers question

2022 · 29 Jul · Shift 1 · Q26
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  5. /2022 · 29 Jul · Shift 1 · Q26

Complex Numbers question

2022 · 29 Jul · Shift 1 · Q26

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If z=2+3iz=2+3 iz=2+3i, then z5+(zˉ)5z^{5}+(\bar{z})^{5}z5+(zˉ)5 is equal to :
  1. A
    244
  2. B
    224
  3. C
    245
  4. D
    265
View written solutionFree

Correct answer: A

  1. Given

    \quad \bar z=2-3i$$

  2. We need to find z5+(zˉ)5.z^5+(\bar z)^5.z5+(zˉ)5.

    Since zˉ\bar zzˉ is the conjugate of zzz, we use the fact: z5+(zˉ)5=2Re⁡(z5).z^5+(\bar z)^5 = 2\operatorname{Re}(z^5).z5+(zˉ)5=2Re(z5).

  3. First compute powers of zzz step-by-step.

    z2=(2+3i)2=4+12i+9i2=4+12i−9=−5+12iz^2=(2+3i)^2=4+12i+9i^2=4+12i-9=-5+12iz2=(2+3i)2=4+12i+9i2=4+12i−9=−5+12i

  4. Then z4=(z2)2=(−5+12i)2=25−120i+144i2=25−120i−144=−119−120iz^4=(z^2)^2=(-5+12i)^2=25-120i+144i^2=25-120i-144=-119-120iz4=(z2)2=(−5+12i)2=25−120i+144i2=25−120i−144=−119−120i

  5. Now compute z5z^5z5: z5=z4⋅z=(−119−120i)(2+3i)z^5=z^4\cdot z=(-119-120i)(2+3i)z5=z4⋅z=(−119−120i)(2+3i)

    Expanding, (−119)(2)+(−119)(3i)+(−120i)(2)+(−120i)(3i)(-119)(2)+(-119)(3i)+(-120i)(2)+(-120i)(3i)(−119)(2)+(−119)(3i)+(−120i)(2)+(−120i)(3i) =−238−357i−240i−360i2=-238-357i-240i-360i^2=−238−357i−240i−360i2

    Since i2=−1i^2=-1i2=−1, z5=−238−597i+360=122−597iz^5=-238-597i+360=122-597iz5=−238−597i+360=122−597i

  6. Therefore, (zˉ)5=122+597i(\bar z)^5=122+597i(zˉ)5=122+597i

  7. Add them: z5+(zˉ)5=(122−597i)+(122+597i)=244z^5+(\bar z)^5=(122-597i)+(122+597i)=244z5+(zˉ)5=(122−597i)+(122+597i)=244

  8. Checking options:

    • A: 244 ✅
    • B: 224
    • C: 245
    • D: 265

So the correct answer is A.

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