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Write out the real and imaginary parts for <math>\log(a_1a_2)</math> then choose the principal branch of log and add <math>2n\pi</math> to get all the possible branch choices. Next do the same for <math>\log(a_1)+\log(a_2)</math> only using two different variables (one for each number) for the possible branches. Adding the results of the two logs together should give a relation between all the variables. Hope this helps --[[User:Rgilhamw|Rgilhamw]] 18:47, 4 October 2009 (UTC) | Write out the real and imaginary parts for <math>\log(a_1a_2)</math> then choose the principal branch of log and add <math>2n\pi</math> to get all the possible branch choices. Next do the same for <math>\log(a_1)+\log(a_2)</math> only using two different variables (one for each number) for the possible branches. Adding the results of the two logs together should give a relation between all the variables. Hope this helps --[[User:Rgilhamw|Rgilhamw]] 18:47, 4 October 2009 (UTC) | ||
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+ | So for 9.3 is it enough just to say <math>arg(a1*a2)=arg(a1)+arg(a2)</math> ?? | ||
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+ | Also Prof Bell, If you do read this could you please post the lecture notes from last class online. | ||
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+ | Thanks --[[User:Kfernan|Kfernan]] 20:12, 4 October 2009 (UTC) |
Revision as of 15:12, 4 October 2009
Homework 5
For problem 9.3, how detailed should our explanation be? Is a mathematical proof required along with our reasoning in words? --Ysuo
Write out the real and imaginary parts for $ \log(a_1a_2) $ then choose the principal branch of log and add $ 2n\pi $ to get all the possible branch choices. Next do the same for $ \log(a_1)+\log(a_2) $ only using two different variables (one for each number) for the possible branches. Adding the results of the two logs together should give a relation between all the variables. Hope this helps --Rgilhamw 18:47, 4 October 2009 (UTC)
So for 9.3 is it enough just to say $ arg(a1*a2)=arg(a1)+arg(a2) $ ??
Also Prof Bell, If you do read this could you please post the lecture notes from last class online.
Thanks --Kfernan 20:12, 4 October 2009 (UTC)