Solve aq=(1-Y/1+Y)^2

$$a q = ( \frac { 1 - Y } { 1 + Y } ) ^ { 2 }$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=\frac{\left(\frac{Y-1}{Y+1}\right)^{2}}{q}\text{, }&Y\neq -1\text{ and }q\neq 0\\a\in \mathrm{C}\text{, }&q=0\text{ and }Y=1\end{matrix}\right.$$

Steps for Solving Linear Equation

To raise \frac{1-Y}{1+Y} to a power, raise both numerator and denominator to the power and then divide.

$$aq=\frac{\left(1-Y\right)^{2}}{\left(1+Y\right)^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-Y\right)^{2}.

$$aq=\frac{1-2Y+Y^{2}}{\left(1+Y\right)^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+Y\right)^{2}.

$$aq=\frac{1-2Y+Y^{2}}{1+2Y+Y^{2}}$$

Multiply both sides of the equation by \left(Y+1\right)^{2}.

$$aq\left(Y+1\right)^{2}=1-2Y+Y^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(Y+1\right)^{2}.

$$aq\left(Y^{2}+2Y+1\right)=1-2Y+Y^{2}$$

Use the distributive property to multiply aq by Y^{2}+2Y+1.

$$aqY^{2}+2aqY+aq=1-2Y+Y^{2}$$

Combine all terms containing a.

$$\left(qY^{2}+2qY+q\right)a=1-2Y+Y^{2}$$

The equation is in standard form.

$$\left(qY^{2}+2Yq+q\right)a=Y^{2}-2Y+1$$

Divide both sides by qY^{2}+2Yq+q.

$$\frac{\left(qY^{2}+2Yq+q\right)a}{qY^{2}+2Yq+q}=\frac{\left(Y-1\right)^{2}}{qY^{2}+2Yq+q}$$

Dividing by qY^{2}+2Yq+q undoes the multiplication by qY^{2}+2Yq+q.

$$a=\frac{\left(Y-1\right)^{2}}{qY^{2}+2Yq+q}$$

Divide \left(Y-1\right)^{2} by qY^{2}+2Yq+q.

$$a=\frac{\left(Y-1\right)^{2}}{q\left(Y+1\right)^{2}}$$

Solve for a

$$\left\{\begin{matrix}a=\frac{\left(\frac{Y-1}{Y+1}\right)^{2}}{q}\text{, }&Y\neq -1\text{ and }q\neq 0\\a\in \mathrm{R}\text{, }&q=0\text{ and }Y=1\end{matrix}\right.$$

Steps for Solving Linear Equation

To raise \frac{1-Y}{1+Y} to a power, raise both numerator and denominator to the power and then divide.

$$aq=\frac{\left(1-Y\right)^{2}}{\left(1+Y\right)^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-Y\right)^{2}.

$$aq=\frac{1-2Y+Y^{2}}{\left(1+Y\right)^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+Y\right)^{2}.

$$aq=\frac{1-2Y+Y^{2}}{1+2Y+Y^{2}}$$

Multiply both sides of the equation by \left(Y+1\right)^{2}.

$$aq\left(Y+1\right)^{2}=1-2Y+Y^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(Y+1\right)^{2}.

$$aq\left(Y^{2}+2Y+1\right)=1-2Y+Y^{2}$$

Use the distributive property to multiply aq by Y^{2}+2Y+1.

$$aqY^{2}+2aqY+aq=1-2Y+Y^{2}$$

Combine all terms containing a.

$$\left(qY^{2}+2qY+q\right)a=1-2Y+Y^{2}$$

The equation is in standard form.

$$\left(qY^{2}+2Yq+q\right)a=Y^{2}-2Y+1$$

Divide both sides by qY^{2}+2Yq+q.

$$\frac{\left(qY^{2}+2Yq+q\right)a}{qY^{2}+2Yq+q}=\frac{\left(Y-1\right)^{2}}{qY^{2}+2Yq+q}$$

Dividing by qY^{2}+2Yq+q undoes the multiplication by qY^{2}+2Yq+q.

$$a=\frac{\left(Y-1\right)^{2}}{qY^{2}+2Yq+q}$$

Divide \left(Y-1\right)^{2} by qY^{2}+2Yq+q.

$$a=\frac{\left(Y-1\right)^{2}}{q\left(Y+1\right)^{2}}$$

Solve for Y (complex solution)

$$\left\{\begin{matrix}\\Y=\frac{-\sqrt{a}\sqrt{q}+1}{\sqrt{a}\sqrt{q}+1}\text{, }&\text{unconditionally}\\Y=-\frac{\sqrt{a}\sqrt{q}+1}{\sqrt{a}\sqrt{q}-1}\text{, }&q=0\text{ or }a\neq \frac{1}{q}\end{matrix}\right.$$

Solve for Y

$$\left\{\begin{matrix}Y=-\frac{\sqrt{aq}+1}{\sqrt{aq}-1}\text{, }&a\geq 0\text{ and }\left(q=0\text{ or }a\neq \frac{1}{q}\right)\text{ and }q\geq 0\\Y=\frac{-\sqrt{aq}+1}{\sqrt{aq}+1}\text{, }&\left(q\geq 0\text{ and }a\geq 0\right)\text{ or }\left(a\leq 0\text{ and }q\leq 0\right)\\Y=-\frac{\left(\sqrt{aq}+1\right)^{2}}{aq-1}\text{, }&a\leq 0\text{ and }\left(q=0\text{ or }a\neq \frac{1}{q}\right)\text{ and }q\leq 0\end{matrix}\right.$$