First, we express the base 1+i in its polar form. The magnitude is r = √(1² + 1²) = √2, and the argument is θ = ((1)/(1)) = (π)/(4). So, 1+i = √2 e^(i(π)/(4)). We also need to consider the general form including 2nπ for the argument, so 1+i = √2 e^(i((π)/(4) + 2nπ)) for integer n.

First, we express the base 1+i in its polar form. The magnitude is r = √(1² + 1²) = √2, and the argument is θ = ((1)/(1)) = (π)/(4). So, 1+i = √2 e^(i(π)/(4)). We also need to consider the general form including 2nπ for the argument, so 1+i = √2 e^(i((π)/(4) + 2nπ)) for integer n.

Substitute the values of w and ln z into the expression w ln z. We have w = 1-i and ln z = ln√2 + i((π)/(4) + 2nπ) = (1)/(2)ln 2 + i((π)/(4) + 2nπ). Multiply these two complex numbers and group the real and imaginary parts.

Substitute the values of w and ln z into the expression w ln z. We have w = 1-i and ln z = ln√2 + i((π)/(4) + 2nπ) = (1)/(2)ln 2 + i((π)/(4) + 2nπ). Multiply these two complex numbers and group the real and imaginary parts.

Find all values of (1+i)^(1-i) in the form a + bi.

Answer: $e^{\left( \frac{1}{2}\ln 2 + \frac{\pi}{4} + 2n\pi \right)} \left[ \cos\left( \frac{\pi}{4} + 2n\pi - \frac{1}{2}\ln 2 \right) + i \sin\left( \frac{\pi}{4} + 2n\pi - \frac{1}{2}\ln 2 \right) \right]$ for $n \in \mathbb{Z}$

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 $r = \sqrt{1^2 + 1^2} = \sqrt{2}$ , and the argument is $\theta = \arctan\left(\frac{1}{1}\right) = \frac{\pi}{4}$ . So, $1+i = \sqrt{2} e^{i\frac{\pi}{4}}$ . We also need to consider the general form including $2n\pi$ for the argument, so $1+i = \sqrt{2} e^{i(\frac{\pi}{4} + 2n\pi)}$ for integer $n$ .

Step 2: Apply the definition of complex exponentiation

We use the general definition of complex exponentiation, which states that $z^w = e^{w \ln z}$ . Here, $z = 1+i$ and $w = 1-i$ . The natural logarithm of a complex number $z = re^{i\theta}$ is $\ln z = \ln r + i\theta$ .

Step 3: Calculate $w \ln z$

Substitute the values of $w$ and $\ln z$ into the expression $w \ln z$ . We have $w = 1-i$ and $\ln z = \ln\sqrt{2} + i\left(\frac{\pi}{4} + 2n\pi\right) = \frac{1}{2}\ln 2 + i\left(\frac{\pi}{4} + 2n\pi\right)$ . Multiply these two complex numbers and group the real and imaginary parts.

Step 4: Substitute into $e^{w \ln z}$

Now we substitute the calculated value of $w \ln z$ back into the exponential form $e^{w \ln z}$ . This gives us the general form of $(1+i)^{(1-i)}$ .

Step 5: Separate into real and imaginary parts

Using Euler's formula, $e^{A+Bi} = e^A (\cos B + i \sin B)$ , we separate the expression into its real and imaginary components. Here, $A = \frac{1}{2}\ln 2 + \frac{\pi}{4} + 2n\pi$ and $B = \frac{\pi}{4} + 2n\pi - \frac{1}{2}\ln 2$ . This gives the final form $a+bi$ .

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