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Показувати елементи керування програвачем
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I used the first method.
This problem is an example of when we DON'T need to put e^i.2pi.k on the lhs. The k and n will simply combine into one integer.
1+i = (√2)e^(iπ/4) so z = ln(1+i) = ln√2 + πi/4 + 2kπi ∀k∈ℤ = ln√2 + (8k+1)πi/4.
I used the first method.
This problem is an example of when we DON'T need to put e^i.2pi.k on the lhs. The k and n will simply combine into one integer.
1+i = (√2)e^(iπ/4) so z = ln(1+i) = ln√2 + πi/4 + 2kπi ∀k∈ℤ = ln√2 + (8k+1)πi/4.