Answer
The standard form of the expression $\frac{1+i}{1-i}$ is $0+i$.
Work Step by Step
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 $0+i$.
