Question

# Let z=(√32+i2)5+(√32−i2)5. If R(z) and I(z) respectively denote the real and imaginary parts of z, then :R(z)>0 and I(z)>0R(z)=−3I(z)=0R(z)<0 and I(z)>0

Solution

## The correct option is C I(z)=0z=(√32+i2)5+(√32−i2)5   ⋯(1) The equation (1) can be written as ⇒z=(eiπ/6)5+(e−iπ/6)5 ⇒z=ei5π/6+e−i5π/6 ⇒z=cos5π6+isin5π6+cos−5π6+isin−5π6 ⇒z=2cos5π6 ⇒z=2(−√32)<0 Im(z)=0 and Re(z)<0

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