Question 4138: Z=1+i√3. Find the smallest positive integer n for which z^n is real and evaluate z^n for this value of n. Show that there is no integral value of n for which z^n is imaginary.
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! Z=1+i√3, note r = √(a^2+b^2) = √(1+3) = 2,
and theta = Arc Tan(b/a) = Arc Tan(√3) = pi/3 (set theta = x)
in polar coordinates, z = r(cos x + i sin x)
= 2(cos pi/3 + i sin pi/3)
By DeMoivre Theorem,
z^n = 2^n(cos pi/3 + i sin pi/3)^n = 2^3(cos n pi/3 + i sin n pi/3).
If z^n is real, then sin npi/3 = 0 , equivalently n pi /3 must
be multiple of pi. We see that when n = 3 , n pi /3 = pi.
Hence,the smallest positive integer n such that z^n is real is 3 and
we have z^3 = 2^3(cos pi + i sin pi) = 8(-1+ i * 0) = -8.
Note that if z^n is imaginary then cos n pi/3 should be 0 and so
n pi/3 must be equal to (2k +1)pi/2 for some integer k.
But n pi/3 = (2k +1)pi/2 implies 2n = 3(2k+1), which is impossible
for any integer because the left side is even while the right hand side is
odd. Hence,z^n cannot be iaginary.
This completes the proof of the two requirements.
Kenny
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