Solve f(1+i/1-i)^x=1

$$f ( \frac { 1 + i } { 1 - i } ) ^ { x } = 1$$

Solve for f

$$f=\frac{1}{i^{x}}$$

Steps for Solving Linear Equation

Multiply both numerator and denominator of \frac{1+i}{1-i} by the complex conjugate of the denominator, 1+i.

$$f\times \left(\frac{\left(1+i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}\right)^{x}=1$$

Do the multiplications in \frac{\left(1+i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}.

$$f\times \left(\frac{2i}{2}\right)^{x}=1$$

Divide 2i by 2 to get i.

$$fi^{x}=1$$

The equation is in standard form.

$$i^{x}f=1$$

Divide both sides by i^{x}.

$$\frac{i^{x}f}{i^{x}}=\frac{1}{i^{x}}$$

Dividing by i^{x} undoes the multiplication by i^{x}.

$$f=\frac{1}{i^{x}}$$

Solve for x

$$x=\frac{2i\ln(f)}{\pi }+4n_{1}$$

$$n_{1}\in \mathrm{Z}$$

$$f\neq 0$$