Engineering
Mathematics
Introduction to Complex Number
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) > 0

R(z) < 0 and I(z) > 0

R(z) = – 3

I(z) = 0

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Solution

(32+i2)5=(eiπb)5=ei5π/6

∴z=2cos5π6=2(−32)=−3