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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)

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Solution The Specific Problem | Comment 5 of 14 |

The specific problem with the false equality re^(it)= r(e^(2PIi))^(t/2PI) is that here e^(it) is "just notation" and is not really the it power of e=2.718... at all.   It just stands for exp(it)=cos(t)+isin(t) where cos(t) and sin(t) are the ordinary real cosine and sine functions of the real angle t (in radians).

There is no identity exp(ab)=(exp(a))^b. The identity that the exponential function satisfies is exp(a+b)=exp(a)exp(b).

(Edited because cut and paste gave this kind of garbage:

Now, z= r*e^(i)= r*(e^(2i))^(/2). Now as we know that e^(2i)=1 we can write z =r*(1)^(/2) z=r. )

 

Edited on August 25, 2006, 10:40 pm
  Posted by Richard on 2006-08-25 22:33:15

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