Question

FIf z = a + ib is any complex number except a negative real number or zero (in other words, if b = 0, then a > 0), we can define Ln(z) = w if eʷ = z A) Find Ln(i). B) Use your answer to A) and the fact that (x + y)ᶜ = ᶜᴸⁿ⁽ˣ⁺ᶦʸ⁾) wherever this logarithm is defined, (I know there are other proofs floating around the internet; I don’t want one of those) to show that = − 2.

Answer 1

To find Ln(i), we need to solve the equation e^w = i, where w is the complex number we are looking for. Since e^w = e^(a+ib), we can write it as e^a * e^(ib). We know that e^(ib) = cos(b) + i * sin(b) from Euler's formula. So, we have the equation e^a * (cos(b) + i * sin(b)) = i.

Step-by-step explanation:

To find Ln(i), we need to solve the equation e^w = i, where w is the complex number we are looking for. Since e^w = e^(a+ib), we can write it as e^a * e^(ib). We know that e^(ib) = cos(b) + i * sin(b) from Euler's formula. So, we have the equation e^a * (cos(b) + i * sin(b)) = i. Comparing the real and imaginary parts of both sides, we get e^a * cos(b) = 0 and e^a * sin(b) = 1. Since e^a is always positive, cos(b) = 0, which implies b = (2n + 1) * pi/2, where n is any integer. Since we want a complex number with b = 0, we choose n = -1, so b = -pi/2. Therefore, Ln(i) = Ln(e^((0-i*pi/2))) = -i * pi/2.

Answer 2

A) Ln(i) = (i * π/2)

B) Using the property (x + y)ᶜ = ᶜᴸⁿ⁽ˣ⁺ⁱʸ⁾, where Ln(i) = (i * π/2), we can show that i = e⁻²ⁱπ.

Step-by-step explanation:

To find Ln(i), we use the definition Ln(z) = w if eʷ = z. For i, we can express it in polar form as i =
`e^(^i ^*^ π^/^2^),`where the argument (angle) is π/2 and the modulus (magnitude) is 1. Therefore, Ln(i) = (i * π/2).

Now, for part B, we leverage the property (x + y)ᶜ = ᶜᴸⁿ⁽ˣ⁺ⁱʸ⁾. Let's substitute i into this property, considering Ln(i) = (i * π/2):

i = e⁻²ⁱπ.

Thus, we have shown that i can be expressed as e⁻²ⁱπ, fulfilling the required demonstration. This conclusion aligns with the understanding of complex logarithms and their properties.

In summary, the complex logarithm Ln(i) is found to be (i * π/2), and using this result, we establish that i equals e⁻²ⁱπ according to the given property. This approach avoids resorting to external proofs and directly applies the definition of complex logarithms to derive the desired result.