find the value of i^i
Answers
Answer:
not sure
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Step-by-step explanation:
What is i to the power of i?
Disclaimer: knowledge of complex numbers and trigonometry is needed to fully appreciate this note.
Some of you may know the definition of ii, the imaginary unit. For those who don't, i=\sqrt{-1}i=
−1
. But what is i^ii
i
? Is it an imaginary number? Perhaps even one step deeper than imaginary, a quaternary number?
In fact, i^ii
i
is a real number! It approximately equals 0.207880.20788.
But how can that be? How can an imaginary number to the power of an imaginary number be real? How did I arrive at that inexplicable answer?
To find the value of i^ii
i
, we'll first need to know Euler's Formula.
Euler's Formula states that e^{ix}=\cos{x}+i\sin{x}e
ix
=cosx+isinx. The right hand side can be shortened to \text{cis }{x}cis x (just notation, don't get confused), and I'll be using this shorthand for brevity.
Note that when x=\dfrac{\pi}{2}x=
2
π
, then \text{cis }{x}=\cos{\dfrac{\pi}{2}}+i\sin{\dfrac{\pi}{2}}\implies\text{cis }{x}=icis x=cos
2
π
+isin
2
π
⟹cis x=i. Therefore, to find i^ii
i
, we can simply take \text{cis }{\dfrac{\pi}{2}}cis
2
π
to the iith power. However, since \text{cis }{\dfrac{\pi}{2}}=e^{i\frac{\pi}{2}}cis
2
π
=e
i
2
π
, then
\begin{aligned}i^i&=(\text{cis }{\dfrac{\pi}{2}})^i\\ &=e^{i^2\frac{\pi}{2}}\\ &=e^{-\frac{\pi}{2}}\\ &\approx 0.20788\end{aligned}
i
i
=(cis
2
π
)
i
=e
i
2
2
π
=e
−
2
π
≈0.20788
and we are done.
However, the astute reader would notice something wrong with my argument. When I said x=\dfrac{\pi}{2}x=
2
π
gives \text{cis }{x}=icis x=i, I was only considering one scenario of the infinite possible xx that satisfy \text{cis }{x}=icis x=i. What about x=\dfrac{5\pi}{2}x=
2
5π
? How about x=\dfrac{-3\pi}{2}x=
2
−3π
? Any xx satisfying x=\dfrac{\pi}{2}+2\pi nx=
2
π
+2πn for some integer nn would work, and every single nn gives a different value for i^ii
i
. Which one is the correct answer?
The short answer is: all of them are equally correct. But how can that be? Shouldn't i^ii
i
, a constant to the power of a constant, yield only one value? One reason that i^ii
i
might not give what is expected is the fact that it is to the power of ii. What does it mean to take an iith power? The fundamental meaning of taking "to the power" breaks down when you consider this.
You can make this weird fact more acceptable to yourself by asking yourself: What are all the possible values of 4^{\frac{1}{2}}4
2
1
? There are, in fact, two values: -2−2 and 22.
Therefore along those lines of reasoning, i^i=0.20788i
i
=0.20788, given by n=0n=0. Also, i^i=111.31778i
i
=111.31778, where n=-1n=−1. i^i=0.00039i
i
=0.00039, where n=1n=1.
But much like the principle value of 4^{\frac{1}{2}}=24
2
1
=2, we also have the principle value of i^ii
i
, which is when n=0n=0. So, finally, we have shown that i^ii
i
does indeed equal 0.207880.20788. \Box□