Why is 1+i = root 2e^(pi/4+2pi k) true?

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The discussion centers on the identity 1+i = √2e^(π/4 + 2πk), exploring the complexities of imaginary numbers and their representation in the complex plane. The original poster expresses confusion over this identity, particularly regarding its implications for cosine and sine values on the unit circle. It is clarified that complex powers can yield multiple values due to the multi-valued nature of complex arguments, as illustrated by the example of i^i. The conversation emphasizes the importance of understanding these nuances when dealing with complex numbers and their properties. Ultimately, the identity holds true when considering the modulus and argument of complex numbers.
The_ArtofScience
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Imaginary numbers have always intrigued me from the very beginning I was taught to believe that some negative root number could have an "i" factored out!

Anyway, I like to know more about this identity if anyone can help me out:

(1+i)^(1+i) = e^ln(root 2)-pi(1/4+2k)e^i(ln(root2)+pi(1/4+2k))

I'm not quite sure as to how 1+i = root 2e^(pi/4+2pi k). I do know that i^i = e^-pi/2 from the identity e^ipi=-1 which gives i = ln(-1)/pi. So my question is why is 1+i = root 2e^(pi/4+2pi k) true? This statement does not seem to make sense to me because it describes both cosine and sine giving 1 as an answer which does not seem to be the case on the unit circle and neither from Euler's identity
 
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1 + i is not root 2.
 
Ha! Sorry found out my own answer

1+i is really just a case of a+ib on the complex xy plane. The modulus can be calculated using (a^2+b^2)^1/2 and the argument can be done using the fact that tangent = b/a

edit:Yes, dx I'm aware of that. I changed it after you made your post
 
The_ArtofScience said:
Imaginary numbers have always intrigued me from the very beginning I was taught to believe that some negative root number could have an "i" factored out!

Anyway, I like to know more about this identity if anyone can help me out:

(1+i)^(1+i) = e^ln(root 2)-pi(1/4+2k)e^i(ln(root2)+pi(1/4+2k))

I'm not quite sure as to how 1+i = root 2e^(pi/4+2pi k). I do know that i^i = e^-pi/2 from the identity e^ipi=-1 which gives i = ln(-1)/pi. So my question is why is 1+i = root 2e^(pi/4+2pi k) true? This statement does not seem to make sense to me because it describes both cosine and sine giving 1 as an answer which does not seem to be the case on the unit circle and neither from Euler's identity

Complex powers are not what they seem. You stated that i^i=e^-pi/2, in fact i^i has infinitely many values:
i^i=e^{ilogi}=e^{i(i(\frac{\pi}{2}+2k\pi))}=e^{-\frac{\pi}{2}}e^{-2k\pi}
Where k is an integer. This 'multi-valuedness' arises from the fact that the argument of a complex number is 'multi-valued', adding an integer multiple of 2pi to the angle a complex number makes with the positive x-axis (real axis) gives you the same number.
Moral of story: Beware of complex powers!

Edit: woops in my haste i forgot to re-read your 3rd line which shows you have taken into account the multivaluedness.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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