problem (-1) ⁽ ᶻ⁺ⁱ ⁾ = ( 1 + i ) Take natural logarithms on both sides. Bring down exponents. ( z + i ) ln (-1) = ln (1 + i ) For the polar form of -1, use the angle π ( 1 + 2 N) where N is an integer. The modulus is 1. -1 = e ⷫ ⁱ ⁽¹⁺²ᴺ⁾ For the polar form of 1 + i use angle π (1/4+2K) = π (8K +1) / 4 and modulus √2. 1 + i = √2 e ⷫ ⁱ ⁽⁸ᴷ⁺¹⁾ᐟ⁴ Replacing and taking the logarithms gives us (z + i) i π (2Ν + 1) = ln√2 + i π (8K + 1)/4 (z + i) i = [ ln√2 + i π (8K + 1)/4 ]/ [π(2N+1)] z + i = [ π (8K + 1)/4 - i ln√2 ]/ [π(2N+1)] z = [ π ( 8K + 1 )/4 - i (ln√2 +1) ]/ [π(2N+1)] z = ( 8K + 1 ) / ( 8N + 4) - i (ln 2 +2) / [π(4N+2)] , where (N, K ∈ ℤ ) answer z = ( 8K + 1 ) / ( 8N + 4) - i (ln 2 +2) / [π (4N+2)] , (N, K ∈ ℤ )
5:52 => It should be (8n+4) in the denominator
Cool!
8n+4 in the denominator.
And in #2 you still need the 2n.pi to get all the solutions.
problem
(-1) ⁽ ᶻ⁺ⁱ ⁾ = ( 1 + i )
Take natural logarithms on both sides. Bring down exponents.
( z + i ) ln (-1) = ln (1 + i )
For the polar form of -1, use the angle π ( 1 + 2 N) where N is an integer. The modulus is 1.
-1 = e ⷫ ⁱ ⁽¹⁺²ᴺ⁾
For the polar form of 1 + i use angle π (1/4+2K) = π (8K +1) / 4 and modulus √2.
1 + i = √2 e ⷫ ⁱ ⁽⁸ᴷ⁺¹⁾ᐟ⁴
Replacing and taking the logarithms gives us
(z + i) i π (2Ν + 1) = ln√2 + i π (8K + 1)/4
(z + i) i = [ ln√2 + i π (8K + 1)/4 ]/ [π(2N+1)]
z + i = [ π (8K + 1)/4 - i ln√2 ]/ [π(2N+1)]
z = [ π ( 8K + 1 )/4 - i (ln√2 +1) ]/ [π(2N+1)]
z = ( 8K + 1 ) / ( 8N + 4) - i (ln 2 +2) / [π(4N+2)]
, where (N, K ∈ ℤ )
answer
z = ( 8K + 1 ) / ( 8N + 4) - i
(ln 2 +2) / [π (4N+2)]
, (N, K ∈ ℤ )
How about a problem including the residue theorem? 🤔
z=-i*ln2/(4k+2)π-i+(8n+1)/(8k+4) is the solution i found but seems a bit off?