Question

Use Problems 27 and 28 to find the following absolute values. If you understand Problems 27 and 28 and equation (5.1), you should be able to do these in your head.

$\left|\frac{1+i}{1-i}\right|$

Short answer

The absolute value of the expression$\left|\frac{1+\mathrm{i}}{1-\mathrm{i}}\right|=1$.

Step-by-step solution

Given information

The given complex number is$\left|\frac{1+\mathrm{i}}{1-\mathrm{i}}\right|$.

Definition of complex numbers

If a and b are real numbers, then a combination of these real numbers with the imaginary number i can be expressed as:

$\mathbf{z}\mathbf{=}\mathbf{a}\mathbf{+}\mathbf{ib}\mathbf{}$

Here z is the complex number.

Use the result concluded in problem 27

Let the complex number be$\left|\mathrm{z}\right|=\left|\frac{1+\mathrm{i}}{1-\mathrm{i}}\right|$.

Use the result concluded in problem 27.

$$\begin{gathered} \left|\mathrm{z}\right|=\left|\frac{{\mathrm{z}}_1}{{\mathrm{z}}_2}\right| \\ \left|\frac{{\mathrm{z}}_1}{{\mathrm{z}}_2}\right|=\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2} \\ \frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\mathrm{R}.......\left(1\right) \end{gathered}$$

Find the modulus of $r_1$and $r_2$ as:

$$\begin{gathered} r_1=\sqrt{1^2+1^2}=\sqrt{2} \\ {r}_2=\sqrt{1^2+1^2}=\sqrt{2} \end{gathered}$$

Put the values in equation (1), we get,

$$\begin{gathered} R=\frac{\sqrt{2}}{\sqrt{2}} \\ R=1 \end{gathered}$$

Hence the absolute value of the expression $\left|\frac{1+\mathrm{i}}{1-\mathrm{i}}\right|=1$ .