Evaluating (1+i)ⁱ | A Very Complex Number?

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(1+i)ⁱ = ?
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⭐ SIMILAR PROBLEMS:

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[Paul-]

The solution given isn’t complete. We started by asking whether the solution was real, and we’re left with another number raised to an imaginary power so we don’t know. The answer given is actually in Re^i(theta) format, with R = e^(-pi/4) (we don’t need the multiples of 2pi) and theta = ln(2) / 2. To get to the a + bi format, use a calculator to find R cos theta and R sin theta. Then we’re done. (The answer is complex. )

[gordonstallings]

correction: It might be nice to include a graph showing the string of solutions spaced e^2pi apart at the angle theta = (ln(2)/2).

[rajeshbuya]

We can as well add the two powers of e in the last step and it results in
e ^ (2kπ - πl4 + i.ln(√2) )

[gordonstallings]

It might be nice to include a graph showing the string of solutions spaced pi apart at the angle theta = (ln(2)/2).

[SuperDeadparrot]

Convert 1+i to sqrt( 2 )*exp( i•pi/4 ) or even better: exp( ln( sqrt( 2 ) ) ) * exp( i•pi/4 ) and then take power of i and get: exp( -pi/4 ) • exp( i•ln( sqrt( 2 ) ) ).

For even more fun, remember that the conversion of 1+i is multi-valued. It makes a factor of exp( 2k•pi ) where k is an integer in the result.

[ManjulaMathew-wbzn]

Since you are so much in to complex numbers I would like to ask a general question.
I get the idea of the complex number system initiated as a result of square rooting a negative number and how it helped making the engineering calculations so much easier.
However if there are no real solutions to an equation does it imply there are complex solutions? Would it be possible to have no real or complex solutions to some ridiculous equations like 1^x=2 or even Sin x =2?
I have seen people solving those but there is no method to check the answer besides reverse calculation which means nothing.
I don’t see any such equation being generated by an engineering application.

[jameeztherandomguy]

I got e^(i * (ln(root2)) - pi/4 + 2*n*pi).

Steps:
(1 + i) = root(2i) = root(2) * e^(pi*i/4)

Raised to the i:

root(2)^i * e^(pi*i/4)^i
or
(e^ln(root2))^i * (e^(pi*i/4))^i

Therefore power rule
e^(i * ln(root2)) * e^(i * pi * i/4)

Simplify
e^(i * ln(root 2)) * e^(-pi/4)


Power rules
e^(i * ln(root 2) - pi/4)

Since 2*n*pi for any n e Z = some multiple of 360, we must add it on

so answer is: exp(i * ln(root(2)) - (pi/4) + 2*n*pi) Right?

[magma]

e^(iln(sqrt(2)-pi/4+2npi) where n in an integer

[cameronspalding]

I would compute
log(1+i) = ln(sqrt(2))+ i*(pi/4+2*pi*n) for some integer n
i*log(1+i) = -(pi/4+2*pi*n) + i*ln(sqrt(2))
(1+i)^i=exp(i*log(1+i))=
exp(-pi/4+2*pi*n) exp(i*ln(sqrt(2)))

[RexxSchneider]

Since ln(√2) = (ln2)/2, we could write (i + i)^i = e^( π(2n - 1/4) + (i.ln2)/2 ), perhaps we could even write it as e^( (π(8n-1) + 2i.ln2) / 4 )

[darkmask]

e^-(π/4+2πn)*cis(½ln2) for some integer n

[samu.bionda]

isn't it easier to call our number z, tranform everything ith exponential notation, take the log on both sides (at this point you should have an equation looking like Converting ipi/4 into its exponential notation you get (pi/4)exp(ipi/2). Combine the exponentials in the equation and ouu get which implies that z=exp(-log(2)pi/8).

[johndoyle]

This math represents a Big Crunch event, where matter is breaking up and more still stable matter is piling on due to inertia generated by gravity.

[SIB]

I love your problems and your explanations.

[cvby]

When you wrote 2npi in the answer's exponent, does n refer to whole numbers too? (-1, -2, etc), asking because the captions werent clear enough... (:

[chrissekely]

What would the answer be in the form "a + bi"?

[fCauneau]

TMHO there are 2 questions in 1:
(1) let the function f(Z)=Z^i, Z in C, evaluate : f(1+i)= ?
(2) solve for Z in C the following equation : Z - (1+i)^i = 0
The video gives the answer for the second formulation, but for the first, we must modify it because the complex Log function (given Z=R e^(i Theta)) has the following restrictions in C:

- R is non zero (complex log is defined for C*)
- 0 <= theta < 2 pi (or any equivalent restriction on the circulation around the R=0 pole)
Otherwise, any of the complex exponentiations and log are no longer *functions* (or applications in C*), as they give more than one value for a single argument. So all of the calculus provided here is correct, but starting with the initial restriction on complex Log, the final solution is given without the '2 n pi' term.

[jmlfa]

The rule (a^b)^c = a^bc … IS NOT VALID FOR IMAGINARY NUMBERS. If it were, then it is easy to show that -1 = 1: Simply replace 1 by e^(2pi*i) in the right part of the equation 1 = sqrt(1).

[antonioorlo]

How about P(a, 2) = a ^ a = 1 + i ; a = ?

[dataweaver]

I haven't watched the video yet.
First, convert the complex number on the base into polar form, where r²=x²+y² and tanθ=y/x. x=1 and y=1, so r²=1²+1²=2 and r=√2. tanθ=y/x=1/1, so θ=π/4+2πn, where n is an arbitrary integer. That means that 1+i=√2·e^{πi/4+2πin}.

Raising that to the power of i multiples the exponent by i, giving us √2·e^{-π/4-2πn}.

(1+i)^i has infinite real solutions, following a geometric progression. Its principal root is √2·e^{-π/4}.