Math425_Practice_Homework_Solutions

# B cos x y cos x cos y sin x sin y solution we recall

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b) cos( x + y ) = cos x cos y - sin x sin y. Solution: We recall that for x,y R , one has: e i ( x + y ) = e ix · e iy . We rewrite both sides by using Euler’s formula to obtain: cos( x + y ) + i sin( x + y ) = (cos x + i sin x ) · (cos y + i sin y ) . It follows that: cos( x + y ) + i sin( x + y ) = (cos x cos y - sin x sin y ) + i (sin x cos y + sin y cos x ) . 1

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2 PRACTICE HOMEWORK FOR MATH 425, SOLUTIONS Claims a) and b) now follow by taking real and imaginary parts of both sides ± . Exercise 3. Find all complex numbers z such that: a) z 6 = 1 . b) z 7 = i. c) Re ( e z ) > 0 . Solution: a) The wanted complex numbers are z k = e 2 πik 6 = e kπi 3 = cos( 3 ) + i sin( 3 ), for k = 0 , 1 ,..., 5. b) Since i = e 2 , we can deduce that the the solutions are given by z k = e 14 + 2 πki 7 , for k = 0 , 1 ,..., 6. c) We write z = re , where r > 0 and θ [0 , 2 π ). In this way, r and θ are uniquely determined from z . Since z = r cos θ + i sin θ , we deduce that: e z = e r (cos θ + i sin θ ) = e r cos θ · e ir sin θ = e r cos θ · ( cos( r sin θ ) + i sin( r sin θ ) ) Since e r cos θ is a positive real number, the condition we need to satisfy is cos( r sin θ ) > 0. An equivalent way to write this is to say that there exists k Z such that: r sin θ
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b cos x y cos x cos y sin x sin y Solution We recall that...

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