+9628 In mathematics, we have positive number,
take the negative of it and we get a negative number.
take the square root of the negative number we get a imaginary number.
add the imaginary number to a real number we get a complex number.
and can we derive anything out of complex number or we just stop there? I feel like the vast ocean of Mathematics is hiding something from me.
~The smartest cookie in the world
+9628 Can someone answer this please...... My curiousity is telling me to take this knowledge ASAP......
+118608 Hi Max,
How about:
\(i^i=real\;number\)
proof:
\(i^i=e^{lni^i}=e^{ilni}\\ \qquad \qquad \text{For complex logs}\\ \qquad \qquad ln(z)=ln|z|+i*arg(z)\\ so\\ ln(i)=ln|i|+i*arg(i)\\ \qquad =ln(1)+i*\frac{\pi}{2}\\ \qquad =i*\frac{\pi}{2}\\~\\ so\\ i^i=e^{lni^i}\\ \:\:\:=e^{i*lni}\\ \:\:\:=e^{i*i*\frac{\pi}{2}}\\ \:\:\:=e^{-1\frac{\pi}{2}}\\ \:\:\:=e^{\frac{-\pi}{2}}\\ \)
So maybe we start going in circles 
+9628 Maybe.
Let's think about \((i^i)^i\) which is \(e^{-i\pi/2}=e^{1/2\cdot\ln(-1)}=\left(e^{\ln(-1)}\right)^{1/2}=\sqrt{-1} = i\)
Then think about \(\left(\left(i^i\right)^i\right)^i\\ \) which is
\(i^i = e^{-i\pi/2}\)
We just go in imaginary -> real -> imaginary -> real -> ......... circles.
We can conclude that:
\(\left(\left(i^i\right)^{.......}\right)^i\text{ with odd number of }i = i\\ \left(\left(i^i\right)^{.......}\right)^i\text{ with even number of }i = e^{-\pi/2}\)
