Solve (1+i)^2+mf(1+i)+n+2=1-i

$$( 1 + i ) ^ { 2 } + m f ( 1 + i ) + n + 2 = 1 - i$$

Solve for f

$$\left\{\begin{matrix}f=-\frac{\left(1-i\right)n+\left(4+2i\right)}{2m}\text{, }&m\neq 0\\f\in \mathrm{C}\text{, }&n=-1-3i\text{ and }m=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Calculate 1+i to the power of 2 and get 2i.

$$2i+mf\left(1+i\right)+n+2=1-i$$

Subtract 2i from both sides.

$$mf\left(1+i\right)+n+2=1-i-2i$$

Subtract 2i from 1-i to get 1-3i.

$$mf\left(1+i\right)+n+2=1-3i$$

Subtract n from both sides.

$$mf\left(1+i\right)+2=1-3i-n$$

Subtract 2 from both sides.

$$mf\left(1+i\right)=1-3i-n-2$$

Do the additions in 1-3i-2.

$$mf\left(1+i\right)=-n-1-3i$$

The equation is in standard form.

$$\left(1+i\right)mf=-1-3i-n$$

Divide both sides by \left(1+i\right)m.

$$\frac{\left(1+i\right)mf}{\left(1+i\right)m}=\frac{-1-3i-n}{\left(1+i\right)m}$$

Dividing by \left(1+i\right)m undoes the multiplication by \left(1+i\right)m.

$$f=\frac{-1-3i-n}{\left(1+i\right)m}$$

Divide -n+\left(-1-3i\right) by \left(1+i\right)m.

$$f=\frac{\left(\frac{1}{2}-\frac{1}{2}i\right)\left(-1-3i-n\right)}{m}$$

Solve for m

$$\left\{\begin{matrix}m=-\frac{\left(1-i\right)n+\left(4+2i\right)}{2f}\text{, }&f\neq 0\\m\in \mathrm{C}\text{, }&n=-1-3i\text{ and }f=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Calculate 1+i to the power of 2 and get 2i.

$$2i+mf\left(1+i\right)+n+2=1-i$$

Subtract 2i from both sides.

$$mf\left(1+i\right)+n+2=1-i-2i$$

Subtract 2i from 1-i to get 1-3i.

$$mf\left(1+i\right)+n+2=1-3i$$

Subtract n from both sides.

$$mf\left(1+i\right)+2=1-3i-n$$

Subtract 2 from both sides.

$$mf\left(1+i\right)=1-3i-n-2$$

Do the additions in 1-3i-2.

$$mf\left(1+i\right)=-n-1-3i$$

The equation is in standard form.

$$\left(1+i\right)fm=-1-3i-n$$

Divide both sides by \left(1+i\right)f.

$$\frac{\left(1+i\right)fm}{\left(1+i\right)f}=\frac{-1-3i-n}{\left(1+i\right)f}$$

Dividing by \left(1+i\right)f undoes the multiplication by \left(1+i\right)f.

$$m=\frac{-1-3i-n}{\left(1+i\right)f}$$

Divide -n+\left(-1-3i\right) by \left(1+i\right)f.

$$m=\frac{\left(\frac{1}{2}-\frac{1}{2}i\right)\left(-1-3i-n\right)}{f}$$