Q1MP - Mathematical Methods in Physical Sciences

Question

Find one or more values of each of the following complex expressions and compare with a computer solution.

${\left(\frac{1+i}{1-i}\right)}^{\mathbf{2718}}$

Step-by-Step Solution

Answer

The value of the complex number is, $z=-1$ .

Step 1: Given Information.

The given expression is, ${\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)}^{2718}$ .

Step 2: Definition of complex series.

The numbers that are presented in the form of $\mathit{a}\mathbf{+}\mathit{i}\mathit{b}$ where, a is real numbers and ' ib ' is an imaginary number called complex numbers.

Example: $3+2i$ .

Step 3: Convert into polar form.

Consider.

$$\begin{gathered} z_1=1+i \\ {z}_2=1-i \end{gathered}$$

 

Write both in polar form.

$$\begin{gathered} z_1=1+i \\ {z}_1=\sqrt{2}{e}^{\left(\mathrm{\pi i}/4\right)} \\ {z}_2=1-i \\ {z}_2=\sqrt{2}{e}^{\left(-\mathrm{\pi i}/4\right)} \end{gathered}$$

Step 4: Substitute the values and solve.

Put both the value in the given question.

$$\begin{gathered} z={\left(\frac{1+i}{1-i}\right)}^{2718} \\ z={\left(\frac{\sqrt{2}{e}^{\left(\mathrm{\pi i}/4\right)}}{\sqrt{2}{e}^{\left(-\mathrm{\pi i}/4\right)}}\right)}^{2718} \\ z={\left[e^{\left(\mathrm{\pi i}/4\right)}\right]}^{2718} \\ z=e^{\left(1359\mathrm{\pi i}\right)} \end{gathered}$$

$$\begin{gathered} z=\cos \left(1359\mathrm{\pi }\right)+\mathrm{i} \sin \left(1359\mathrm{\pi }\right) \\ \mathrm{z}=\cos (1359\mathrm{\pi }-2\mathrm{n\pi })+\mathrm{i} \sin (1359\mathrm{\pi }-2\mathrm{n\pi }) \\ \mathrm{z}=\cos (1359\mathrm{\pi }-1358\mathrm{\pi })+\mathrm{i} \sin (1359\mathrm{\pi }-1358\mathrm{\pi }) \\ \mathrm{z}=\cos \left(\mathrm{\pi }\right)+\mathrm{i} \sin \left(\mathrm{\pi }\right) \end{gathered}$$

 

$z=-1$

 

Hence the value is found to be ${\left(\frac{1+i}{1-i}\right)}^{2718}=-1$