Answer:
the domain of Log (4 + i - 1) (principal value) for it to be differentiable is the complex number 3 + i.
Explanation:
To find the domain of the logarithm function Log (4 + i - 1), we need to consider the values for which the function is defined and differentiable.
The principal value of the logarithm function is defined as Log(z) = ln|z| + iArg(z), where ln|z| represents the natural logarithm of the magnitude of z and iArg(z) represents the principal argument of z.
In this case, the expression 4 + i - 1 simplifies to 3 + i. To determine the domain of this logarithm function, we need to consider the values for which 3 + i is defined.
The magnitude of 3 + i can be found using the Pythagorean theorem: |3 + i| = sqrt((3^2) + (1^2)) = sqrt(10).
The principal argument of 3 + i can be found using trigonometry. The angle θ between the positive real axis and the vector (3, 1) is given by tan(θ) = 1/3. Solving for θ, we find θ = arctan(1/3).
Therefore, the principal value of Log (3 + i) is ln|3 + i| + iArg(3 + i) = ln(sqrt(10)) + i(arctan(1/3)).
To ensure differentiability, the principal argument should be within the range (-π, π]. Therefore, arctan(1/3) should be within this range.
Since arctan(1/3) is approximately 0.32175, it lies within the range (-π, π], and Log (3 + i) is differentiable.