Solve P=R(1-(1+i)^-n/i)*(1+i)

$$P = R ( \frac { 1 - ( 1 + i ) ^ { - n } } { i } ) * ( 1 + i )$$

Solve for R

$$\left\{\begin{matrix}R=\frac{\left(-\frac{1}{2}+\frac{1}{2}i\right)P\left(1+i\right)^{n}}{i\left(1+i\right)^{n}-i}\text{, }&\nexists n_{2}\in \mathrm{Z}\text{ : }n=\frac{i\times 8\pi n_{2}}{2\ln(2)+\pi i}\\R\in \mathrm{C}\text{, }&P=0\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }n=\frac{8\pi n_{1}i}{2\ln(2)+\pi i}\end{matrix}\right.$$

Steps for Solving Linear Equation

Divide each term of 1-\left(1+i\right)^{-n} by i to get \frac{1}{i}-\frac{\left(1+i\right)^{-n}}{i}.

$$P=R\left(\frac{1}{i}-\frac{\left(1+i\right)^{-n}}{i}\right)\left(1+i\right)$$

Multiply both numerator and denominator of \frac{1}{i} by imaginary unit i.

$$P=R\left(\frac{i}{-1}-\frac{\left(1+i\right)^{-n}}{i}\right)\left(1+i\right)$$

Divide i by -1 to get -i.

$$P=R\left(-i-\frac{\left(1+i\right)^{-n}}{i}\right)\left(1+i\right)$$

Use the distributive property to multiply R by -i-\frac{\left(1+i\right)^{-n}}{i}.

$$P=\left(-iR+R\left(-\frac{\left(1+i\right)^{-n}}{i}\right)\right)\left(1+i\right)$$

Use the distributive property to multiply -iR+R\left(-\frac{\left(1+i\right)^{-n}}{i}\right) by 1+i.

$$P=\left(1-i\right)R+\left(1+i\right)R\left(-\frac{\left(1+i\right)^{-n}}{i}\right)$$

Swap sides so that all variable terms are on the left hand side.

$$\left(1-i\right)R+\left(1+i\right)R\left(-\frac{\left(1+i\right)^{-n}}{i}\right)=P$$

Multiply 1+i and -1 to get -1-i.

$$\left(1-i\right)R+\left(-1-i\right)R\times \frac{\left(1+i\right)^{-n}}{i}=P$$

Combine all terms containing R.

$$\left(1-i+\left(-1-i\right)\times \frac{\left(1+i\right)^{-n}}{i}\right)R=P$$

The equation is in standard form.

$$\left(\frac{-1+i}{\left(1+i\right)^{n}}+\left(1-i\right)\right)R=P$$

Divide both sides by 1-i+\left(-1+i\right)\left(1+i\right)^{-n}.

$$\frac{\left(\frac{-1+i}{\left(1+i\right)^{n}}+\left(1-i\right)\right)R}{\frac{-1+i}{\left(1+i\right)^{n}}+\left(1-i\right)}=\frac{P}{\frac{-1+i}{\left(1+i\right)^{n}}+\left(1-i\right)}$$

Dividing by 1-i+\left(-1+i\right)\left(1+i\right)^{-n} undoes the multiplication by 1-i+\left(-1+i\right)\left(1+i\right)^{-n}.

$$R=\frac{P}{\frac{-1+i}{\left(1+i\right)^{n}}+\left(1-i\right)}$$

Divide P by 1-i+\left(-1+i\right)\left(1+i\right)^{-n}.

$$R=\frac{P\left(1+i\right)^{n}}{\left(1-i\right)\left(1+i\right)^{n}+\left(-1+i\right)}$$

Solve for P

$$P=\frac{\left(-1+i\right)R}{\left(1+i\right)^{n}}+\left(1-i\right)R$$