Q19P - Mathematical Methods in Physical Sciences

Question

Verify the formulas in Problems 17 to 24.

$\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\mathbf{ }\mathit{z}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{i}}\mathbf{ }\frac{\mathbf{l}\mathbf{n}\mathbf{1}\mathbf{+}\mathbf{i}\mathbf{z}}{\mathbf{1}\mathbf{-}\mathbf{i}\mathbf{z}}$

Step-by-Step Solution

Answer

The formula is verified $arctan z=\frac{1}{2\mathrm{i}} \frac{\ln 1+\mathrm{iz}}{1-\mathrm{iz}}$.

Step 1: Given Information

Given formula is $arctan z=\frac{1}{2\mathrm{i}} \frac{\ln 1+\mathrm{iz}}{1-\mathrm{iz}}$

Step 2: Definition of Trigonometric equation.

A trigonometric equation is one that has one or more trigonometric ratios with unknown angles.

Step 3: Use exponential form to expand the equation

Given the function $arctan z=\frac{1}{2\mathrm{i}} \frac{\ln 1+\mathrm{iz}}{1-\mathrm{iz}}$.

Write the exponential form of the $\tan ( w)=z.$.

                        $$\begin{gathered} \tan \left(w\right)=\frac{\sin \left(w\right)}{\cos \left(w\right)} \\ =\frac{exp\left(wi\right)-exp\left(-wi\right)}{i\left[exp\left(wi\right)+exp\left(-wi\right)\right]} \end{gathered}$$

  $=\frac{exp\left(wi\right)-exp\left(-wi\right)}{i\left[exp\left(wi\right)+exp\left(-wi\right)\right]}=z .....\left(1\right)$

Rewrite the equation (1).

$$\begin{gathered} e^{\left(wi\right)}-e^{\left(-wi\right)}=z\left[e^{\left(wi\right)}+e^{\left(-wi\right)}\right] \\ {e}^{\left(2wi\right)}-1=iz\left[e^{\left(2wi\right)}+1\right] \\ {e}^{\left(2wi\right)}-1=iz{e}^{\left(2wi\right)}-1+iz \\ {e}^{\left(2wi\right)}\left[1-iz\right]=\left[1+iz\right] \\ e^{\left(2wi\right)}=\frac{1+iz}{1-iz} .......\left(2\right) \end{gathered}$$

Simplify equation (2).

$$\begin{gathered} 2wi=\ln \left(\frac{1+iz}{1-iz}\right) \\ w=\frac{\ln \left(\frac{1+iz}{1-iz}\right)}{2i} \\ w=\frac{1}{2i}\ln \frac{1+iz}{1-iz} \end{gathered}$$

 

Replace value of by $w=arc\tan ( z)$.

$arctan z=\frac{1}{2i}\ln \frac{1+iz}{1-iz}$

 

Hence the formula is verified