Q7P - Mathematical Methods in Physical Sciences

Question

Find each of the following in rectangular form $\mathit{x}\mathbf{+}\mathit{i}\mathit{y}$ and check your results by computer. Remember to save time by doing as much as you can in your head.

$\mathbf{tan}\mathbf{\left(}\mathit{i}\mathbf{ln}\mathbf{2}\mathbf{\right)}$

Step-by-Step Solution

Answer

The rectangular form of the given question $\tan \left(i\ln 2\right)=\frac{3i}{5}.$ .

Step 1: Given Information.

The given expression is $\tan \left(i\ln 2\right)$ .

Step 2: Meaning of rectangular form.

Representing the complex number in rectangular form means writing the given complex number in the form of $\mathit{x}\mathbf{+}\mathit{i}\mathit{y}$, in which x is the real part and y is the imaginary part.

Step 3: Simplify.

Use the complex formula $(\sin \theta ,\cos \theta )=\left(\frac{e^{i\theta }-e^{-i\theta }}{2i},\frac{e^{i\theta }+e^{-i\theta }}{2}\right)$ to rewrite the above expression.

$$\begin{gathered} \tan \theta =\frac{\sin \theta }{\cos \theta } \\ \tan \theta =\frac{e^{i\theta }-e^{-i\theta }}{2i}\left(\frac{2}{e^{i\theta }+e^{-i\theta }}\right) \\ \tan \theta =\frac{e^{i\theta }-e^{-i\theta }}{i\left(e^{i\theta }+e^{-i\theta }\right)} \end{gathered}$$

Step 4: Convert it into rectangular form.

Put the derived expression in the question and solve further.

 $$\begin{gathered} \tan \left(i\ln 2\right)=\frac{e^{i(i\ln 2)}-e^{-i(i\ln 2)}}{i\left(e^{i(i\ln 2)}+e^{-i(i\ln 2)}\right)} \\ \tan \left(i\ln 2\right)=\frac{e^{i^2\ln 2}-e^{-i^2\ln 2}}{i\left(e^{i(i\ln 2)}+e^{-i(i\ln 2)}\right)} \\ \tan \left(i\ln 2\right)=\frac{e^{i^2\ln 2}-e^{-i^2\ln 2}}{i\left(e^{i^2\ln 2}+e^{-i^2\ln 2}\right)} \\ \tan \left(i\ln 2\right)=\frac{e^{-\ln 2}-e^{\ln 2}}{i\left(e^{-\ln 2}+e^{\ln 2}\right)} \end{gathered}$$

Step 5: Solve the mathematical expression.

Write the derived expression and solve further.

$$\begin{gathered} \tan \left(i\ln 2\right)=\frac{e^{\ln 2^{-1}}-e^{\ln 2}}{i\left(e^{\ln 2^{-1}}-e^{\ln 2}\right)} \\ \tan \left(i\ln 2\right)=\frac{2^{-1}-2}{i(2^{-1}+2)} \\ \tan \left(i\ln 2\right)=\frac{(1-4)2}{2i\left(1+4\right)} \\ \tan \left(i\ln 2\right)=\frac{-3}{5i} \end{gathered}$$

 

Multiply and divide with i .

$$\begin{gathered} \tan \left(i\ln 2\right)=\frac{-3i}{5i^2} \\ \tan \left(i\ln 2\right)=\frac{-3i}{-5} \\ \tan \left(i\ln 2\right)=\frac{3i}{5} \end{gathered}$$

 

Therefore, the rectangular form of $\tan \left(i\ln 2\right)$ is $\frac{3i}{5}$ .