## Precalculus (6th Edition) Blitzer

The standard form of the expression $\frac{1+i}{1-i}$ is $0+i$.
Consider the expression, $\frac{1+i}{1-i}$ Since, the imaginary part is in the denominator, multiply the numerator and denominator by the complex conjugate of the denominator; for the complex number $\left( 1-i \right)$, its complex conjugate is $\left( 1+i \right)$. Multiply the expression by $\frac{\left( 1+i \right)}{\left( 1+i \right)}$. \begin{align} & \frac{1+i}{1-i}=\frac{1+i}{1-i}\cdot \frac{1+i}{1+i} \\ & =\frac{\left( 1+i \right)\left( 1+i \right)}{\left( 1-i \right)\left( 1+i \right)} \end{align} The product of the complex number $\left( a+bi \right)$ and its complex conjugate $\left( a-bi \right)$ results in a real number: $\left( a+bi \right)\left( a-bi \right)={{a}^{2}}+{{b}^{2}}$. Therefore, \begin{align} & \frac{\left( 1+i \right)\left( 1+i \right)}{\left( 1-i \right)\left( 1+i \right)}=\frac{\left( 1+i \right)\left( 1+i \right)}{{{1}^{2}}+{{1}^{2}}} \\ & =\frac{\left( 1+i \right)\left( 1+i \right)}{2} \\ & =\frac{{{\left( 1+i \right)}^{2}}}{2} \end{align} Apply the formula for square of the sum of two numbers $\frac{{{\left( 1+i \right)}^{2}}}{2}=\frac{1+2i+{{i}^{2}}}{2}$ As ${{i}^{2}}=-1$ Therefore, \begin{align} & \frac{1+2i+{{i}^{2}}}{2}=\frac{1-1+2i}{2} \\ & =\frac{2i}{2} \end{align} So, $\frac{1+i}{1-i}=\frac{2i}{2}$ Simplify the above expression. Thus, \begin{align} & \frac{1+i}{1-i}=\frac{2i}{2} \\ & =i \end{align} Hence, the standard form of the expression $\frac{1+i}{1-i}$ is $0+i$.