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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: ( Back to comment list | You must be logged in to post comments.)
a possible solution | Comment 13 of 14 |
exp^(ab)=(exp^a)6b does not apply to complex is only for real numbers.if it apply to complex no then say b is complex

(exp^(a))^b=(cos a +sin a)^b=cos ab + sin ab
quantity ab is complex, so  cosine or sine of it is meaningless

exp^(θ*i)=cos θ + i sin θ   is not equal to
cos (θ*i)  + sin (θ*i)  as  your  reasoning  apply

  Posted by zohaib ahmed on 2007-03-05 11:57:32
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