Evaluating (1+i)ⁱ | A Very Complex Number?
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- Опубликовано: 18 апр 2023
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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. )
We can as well add the two powers of e in the last step and it results in
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).
It might be nice to include a graph showing the string of solutions spaced pi apart at the angle theta = (ln(2)/2).
Nice work!
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 ) ) ).
I love your problems and your explanations.
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... (:
Since you are so much in to complex numbers I would like to ask a general question.
e^(iln(sqrt(2)-pi/4+2npi) where n in an integer
I got e^(i * (ln(root2)) - pi/4 + 2*n*pi).
I would compute
very nice
I'm wondering if Euler's formula is valid for complex powers? Meaning raised to the power n; where n is complex
The modulus for the final answer has a "+2kπ" in the exponent. What does that mean? How can there be an infinite number of moduli?
Nice video syber
TMHO there are 2 questions in 1:
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 logz=(ipi/4)exp(ipi/2)log(sqrt2)). Converting ipi/4 into its exponential notation you get (pi/4)exp(ipi/2). Combine the exponentials in the equation and ouu get logz=pi/4(exp(ipi))log(sqrt2)=-log(2)pi/8 which implies that z=exp(-log(2)pi/8).
e^-(π/4+2πn)*cis(½ln2) for some integer n
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 )