Complex NumbersHard
Question
If z = 
, then
, then Options
A.Re (z) = 0
B.Im (z) = 0
C.Re (z) > 0, Im (z) > 0
D.Re (z) > 0, Im (z) > 0
Solution
Given, Z =
Now,
and,
∴ z = (-ω)5 + (iω2)5 = - iω2 + iω
= i(ω - ω2) = i(i√3) = - √3
Re (z) < 0 and Im (z) =0
Alternate Solution
We know z +
= 2Re (z)
If
, then
z is purely real. ie Im (z) = 0
Now,
and,
∴ z = (-ω)5 + (iω2)5 = - iω2 + iω
= i(ω - ω2) = i(i√3) = - √3
Re (z) < 0 and Im (z) =0
Alternate Solution
We know z +
= 2Re (z)If
, then z is purely real. ie Im (z) = 0
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