Q7P - Mathematical Methods in Physical Sciences

Question

Evaluate each of the following in x+iy form, and compare with a computer solution.

$\mathit{l}\mathit{n}\mathbf{\left(}\frac{\mathbf{ }\mathbf{1}\mathbf{+}\mathbf{i}}{\mathbf{1}\mathbf{-}\mathbf{i}}\mathbf{\right)}$

Step-by-Step Solution

Answer

The form of the given equation $ln\left(\frac{ 1+\mathrm{i}}{1-\mathrm{i}}\right)$ is,$\left(\frac{\pi }{2}+2n\pi \right)i.$ .

 

Step 1: Given Information.

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

Step 2: Meaning of rectangular form.

Represent the complex number in rectangular form means writing the given complex number in the form of x+iy in which x is the real part and y is the imaginary part.

Step 3: Convert in polar form.

Consider $z_1=(1-i)$

And $z_1=(1+i)$.

 

Write the polar form of the number.

$z_1=r_1{e}^{\left(i\theta \right)}$ 

 

The angle is located in fourth quadrant so the angle must be accordingly.

$$\begin{gathered} r_1=\left|1-i\right| \\ {r}_1=\sqrt{2} \\ {\theta }_1=-\frac{\pi }{4} \end{gathered}$$

 

Put the values in the polar form.

$z=\sqrt{2}{e}^{\left(-\frac{\pi }{4}i\right)}$

Step 4: Convert in polar form.

Write the polar form of the number.

$z\_2=r_2 e^{\left(i\theta \_2\right)}$

 

The angle is located in first quadrant so the angle must be accordingly.

$$\begin{gathered} r_1=\left|1-i\right| \\ {r}_1=\sqrt{2} \\ {\theta }_1=-\frac{\pi }{4} \end{gathered}$$

Put the values in the polar form.

$z=\sqrt{2}{e}^{\left(-\frac{\pi }{4}i\right)}$

 

Put the values in the expression,$z=\frac{z_2}{z_1}$ .

$z=e^{\left(\frac{\pi }{2}\right)i}$

 

Step 5: Write in the form of .

Convert the polar form into the rectangular form.

$$\begin{gathered} w=ln\left(r{e}^{\theta }i\right) \\ w=ln\left(1\right)+\left(\frac{\pi }{2}+2n\pi \right)i \\ w=ln\left(1\right)+\left(\frac{\pi }{2}+2n\pi \right)i \end{gathered}$$

  $w=ln\left(1\right)+\left(\frac{\pi }{2}+2n\pi \right)i$  where $n=0,\pm 1,\pm 2,\pm 3,\dots$

 

Therefore, the form of the given equation $ln\left(\frac{ 1+\mathrm{i}}{1-\mathrm{i}}\right)$ is,$w=ln\left(1\right)+\left(\frac{\pi }{2}+2n\pi \right)i$ .