Need help with question! (1 Viewer)

bobmcbob365

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Here's the method for the alternative method.

Lol, just keep in mind in the future that if a question provides you with information about a complex number that is purely real or imaginary, you can equate the real/imaginary parts, or alternatively use the geometric approach.

But of course, it depends on the question. Sometimes the first method is easier and faster, and sometimes its the second. It's all relative. =p
 

HeroicPandas

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Oh ok so then we get 2^n (sin n pi/3) = 0 right? then solve that somehow?
YES like that, i made lots of typos inside the sine i think xD

using a.b=0 where a=0 b=0

2^n =/= 0 (its an exponential)

then sin( n pi/3)) = 0
sin (n pi/3) = sin(0)

n pi/3 = 0 + k2pi OR n pi/3 = pi + k2pi <------ use of general solutions/ASTC/fundamental formula
n = 0+k2pi (useless) OR n =3 + k6, therefore the smallet positive integer n for which z^n is real is 3 (from bob's working out)
 

HeroicPandas

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Here's the method for the alternative method.

Lol, just keep in mind in the future that if a question provides you with information about a complex number that is purely real or imaginary, you can equate the real/imaginary parts, or alternatively use the geometric approach.

But of course, it depends on the question. Sometimes the first method is easier and faster, and sometimes its the second. It's all relative. =p
Yep, the boss posted up a clearer version, nice job! :D
 

bobmcbob365

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lol no problem. If it continues to produce an error, I can send it to your email if you want. Although, I must warn, that these solutions are really dodgy sometimes. i.e. they skip steps and usually provides unsatisfactory working out, in terms of marking criteria and also uses really dodgy methods.
 

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