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Are all complex numbers real? (Posted on 2006-08-24) Difficulty: 3 of 5
Let's "prove" that every complex number z is real.

If z=0 it's obvious. For all other complex numbers z=r*e^(θi), where r is a real number, and i=√-1.

Now, z= r*e^(θi)= r*(e^(2πi))^(θ/2π). Now as we know that e^(2πi)=1 we can write z =r*(1)^(θ/2π) → z=r.

What's wrong with this?

No Solution Yet Submitted by atheron    
Rating: 3.3333 (3 votes)

Comments: ( You must be logged in to post comments.)
  Subject Author Date
a possible solutionzohaib ahmed2007-03-05 11:57:52
a possible solutionzohaib ahmed2007-03-05 11:57:32
SolutionDanny2007-01-06 22:29:02
No Subjectpopa2006-10-13 10:47:26
possible solutionvivek2006-09-22 14:15:10
re(3): The Specific Problem (fleshed out a bit more)JLo2006-08-28 09:59:09
re(2): The Specific Problem (fleshed out a bit more)Richard2006-08-27 17:42:25
re(2): The Specific Problem (fleshed out a bit more)Richard2006-08-26 22:45:33
Solutionre: The Specific Problem (fleshed out a bit more)JLo2006-08-26 19:11:51
SolutionThe Specific ProblemRichard2006-08-25 22:33:15
I think this is wrong because...JLo2006-08-24 19:24:54
re(2): The General ProblemRichard2006-08-24 17:32:19
Some Thoughtsre: The General ProblemFederico Kereki2006-08-24 16:27:36
Hints/TipsThe General ProblemRichard2006-08-24 15:54:55
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