Verified Solution • 2025–26 Session moderate

QJEEM19Ma7268: JEE Main 2019 Mathematics - Class 11 JEEM19Ma7268 Solution: Let If R(z) and 1(z) respectively denote the real and imaginary …

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}. If R(z) and 1(z) respectively denote the real and imaginary parts of z, then :

Multiple Choice Options

  • A R(z) = −- 3
  • B R(z) < 0 and I(z) > 0
  • C I(z) = 0
  • D R(z) > 0 and I(z) > 0

Step-by-Step Verified Solution

z=(3+i2)5+(3−i2)5z = {\left( {{{\sqrt 3 + i} \over 2}} \right)^5} + {\left( {{{\sqrt 3 - i} \over 2}} \right)^5}

z=(eiπ/6)5+(e−iπ/6)5z = {\left( {{e^{i\pi /6}}} \right)^5} + {\left( {{e^{ - i\pi /6}}} \right)^5}

=ei5π/6+e−i5π/6= {e^{i5\pi /6}} + {e^{ - i5\pi /6}}

=cos⁡5π6+isin⁡5π6+cos⁡(−5π6)+isin⁡(−5π6)= \cos {{5\pi } \over 6} + i{{\sin 5\pi } \over 6} + \cos \left( {{{ - 5\pi } \over 6}} \right) + i\sin \left( {{{ - 5\pi } \over 6}} \right)

= 2\cos {{5\pi } \over 6} &lt; 0

I(z)=0{\rm I}(z) = 0 and {\mathop{\rm Re}\nolimits} (z) &lt; 0

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