Find all values of (1+i)^(1-i) in the form a + bi.
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Step-by-Step Solution
Step 1: Express the base in polar form
First, we express the base 1+i in its polar form. The magnitude is r=12+12=2, and the argument is θ=arctan(11)=4π. So, 1+i=2ei4π. We also need to consider the general form including 2nπ for the argument, so 1+i=2ei(4π+2nπ) for integer n.
Step 2: Apply the definition of complex exponentiation
We use the general definition of complex exponentiation, which states that zw=ewlnz. Here, z=1+i and w=1−i. The natural logarithm of a complex number z=reiθ is lnz=lnr+iθ.
Step 3: Calculate wlnz
Substitute the values of w and lnz into the expression wlnz. We have w=1−i and lnz=ln2+i(4π+2nπ)=21ln2+i(4π+2nπ). Multiply these two complex numbers and group the real and imaginary parts.
Step 4: Substitute into ewlnz
Now we substitute the calculated value of wlnz back into the exponential form ewlnz. This gives us the general form of (1+i)(1−i).
Step 5: Separate into real and imaginary parts
Using Euler's formula, eA+Bi=eA(cosB+isinB), we separate the expression into its real and imaginary components. Here, A=21ln2+4π+2nπ and B=4π+2nπ−21ln2. This gives the final form a+bi.