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there are infinitely-many values, not just one value. Let x = i^(1/i) --> ln(x) = ln(i)/i. Now, i = exp[i*(pi/2 + 2*pi*n)] where n = any integer (negative, 0, positive). --> ln(i) = i*(pi/2 + 2*pi*n) --> ln(x) = pi/2 + 2*pi*n --> x = exp[pi/2 + 2*pi*n] where n = any integer. NOTE: 1. learn polar form. 2. recognize that polar angle theta is not unique. it is periodic in 2*pi.
After expressing the quantity of interest as (i)^(1/i), a simple way to proceed is with Euler's formula, e^(ix) = cosx + i*sinx. Note that e^(i*pi/2) = 0+i*1 = i, so (i)^1/i = [e^(i*pi/2)]^(1/i) = e^(pi/2), the desired result. As another commenter notes, there are an infinite number of repeating solutions because the cosine and sine terms repeat at integral increments of 2*pi, so i=e^[(4n+1)*(pi/2)] for all integer values of n. The solution initially worked out is the one corresponding to n=0.
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Nice solution
thank you so much
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Subscribed
okay
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@AsadInternationalAcademy noted
there are infinitely-many values, not just one value.
Let x = i^(1/i) --> ln(x) = ln(i)/i. Now, i = exp[i*(pi/2 + 2*pi*n)] where n = any integer (negative, 0, positive).
--> ln(i) = i*(pi/2 + 2*pi*n) --> ln(x) = pi/2 + 2*pi*n --> x = exp[pi/2 + 2*pi*n] where n = any integer.
NOTE: 1. learn polar form. 2. recognize that polar angle theta is not unique. it is periodic in 2*pi.
thanks
After expressing the quantity of interest as (i)^(1/i), a simple way to proceed is with Euler's formula, e^(ix) = cosx + i*sinx. Note that e^(i*pi/2) = 0+i*1 = i, so (i)^1/i = [e^(i*pi/2)]^(1/i) = e^(pi/2), the desired result. As another commenter notes, there are an infinite number of repeating solutions because the cosine and sine terms repeat at integral increments of 2*pi, so i=e^[(4n+1)*(pi/2)] for all integer values of n. The solution initially worked out is the one corresponding to n=0.
nice idea, thanks
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Anything to the power of i has infinite results since x^i makes a rotation on the cartesian plane
okay
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hi how do i know i^i = e^-pi/2
i = e^((pi/2)*i) by converting to polar form of an exponential.
so i ^ i = (e^((pi/2)*i))^i = e^((pi/2)*i^2) = e^(-pi/2)
@ ah okay thanks
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thanks