hw2sol - Complex Analysis Spring 2001 Homework II Solutions...

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Unformatted text preview: Complex Analysis Spring 2001 Homework II Solutions 1. With π 2 defined to be the least positive zero of cos t , we established in class Wednesday that t → e it was onto the first quadrant of the unit circle in C . Use proven properties of the complex exponential to prove that cos( π- t ) =- cos t, cos( π + t ) =- cos t , sin( π- t ) = sin t , and sin( π + t ) =- sin t and use these to finish the proof that e it is onto. It was proved that for t ∈ [0 , π 2 ], cos t is decreasing with cos 0 = 1 and cos π 2 = 0 while sin t is increasing, with sin 0 = 0 and sin π 2 = 1 , so e i π 2 = i. Thus e iπ =- 1 . Now e ( π- t ) i = e πi e- ti =- e- it =- 1 e it =- 1 cos t + i sin t =- cos t + i sin t so cos( π- t ) =- cos t and sin( π- t ) = sin t. Also e ( π + t ) i = e iπ e it =- cos t- i sin t, so cos( π + t ) =- cos t and sin( π + t ) =- sin t. From the first relation we see that the values of e it for t ∈ [ π 2 , π ] are reflections of values for t ∈ [0 , π 2 ] and so the part of the unit circle in the 2nd quadrant is in the image of the exponential. The second relation show that the values for t ∈ [ π, 2 π ] are reflections in the origin of values for t ∈ [0 , π ] and hence the part of the unit circle in the 3rd and 4th quadrants is also in the image of e it ....
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This note was uploaded on 10/22/2009 for the course MATH 552 taught by Professor Snider during the Spring '01 term at USC.

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hw2sol - Complex Analysis Spring 2001 Homework II Solutions...

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