TEST 1 SOLUTIONS–MATH 1102

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/10
October 10, 2012 4
7. Compute the standard form of the complex number e−iπ/3 . Your final
answer should not contain any trigonometric functions.
Solution: By definition
e −iπ/3 = e i(−π/3) = cos(−π/3) + i sin(−π/3) = 1/2 − ( √ 3/2)i
Grading: 2 for definition of e i(−π/3) . 2 for trigonometric identity.
8. Prove by induction on n that
( √ 3 + i) n = 2 n (cos(nπ/6) + i sin(nπ/6))
for all integers n ≥ 1. Do not use the fact that (e iθ ) n = e inθ . You may
use trigonometric identities without proof.
Solution: Base case. Suppose n = 1. Then
( √ 3 + i) 1 = √ 3 + i = 2( √ 3/2 + (1/2)i) = 2(cos((1)π/6) + i sin((1)π/6))
and we are done.
Now assume that ( √ 3 + i) n = 2 n (cos(nπ/6) + i sin(nπ/6)).
For the induction step we must prove
( √ 3 + i) n+1 = 2 n+1 (cos((n + 1)π/6) + i sin((n + 1)π/6)).
( √ 3 + i) n+1 = ( √ 3 + i) n ( √ 3 + i)
= (2 n (cos(nπ/6) + i sin(nπ/6))) (2(cos(π/6) + i sin(π/6)))
Here we have used the induction assumption and the base case. We continue
with
= 2 n+1 (cos(nπ/6) + i sin(nπ/6))(cos(π/6) + i sin(π/6))
= 2 n+1 (cos(nπ/6) cos(π/6) − sin(nπ/6) sin(π/6)
+i(cos(nπ/6) sin(π/6) + sin(nπ/6) cos(π/6)))
= 2 n+1 (cos((n + 1)π/6) + i sin((n + 1)π/6)).
We used the identities for adding angles in cos and sin in the final step.
Grading: 2 for proof of base case. 1 for induction assumption. 1 for
statement of induction step. 2 for correct use of induction assumption in
induction step. 4 for remaining proof of induction step.