Q15P - Mathematical Methods in Physical Sciences

Question

Find each of the following in the x + iy form and compare a computer solution.

arctan(2 + i )

Step-by-Step Solution

Answer

The x + iy form of the given equation $\arctan (2+i) \mathrm{is} \frac{i \ln \left(2\right)}{4}+\left(\frac{3\mathrm{\pi }}{8}\pm n\mathrm{\pi }\right)$is 

Step 1: Given Information.

The given expression is $\arctan (2+i).$

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 into polar form.

Consider the complex number $z=\arctan (2+i)$.

Rewrite the above expression.

$\mathrm{tanhz}=2+i$

 

Write the formula for $\tanh \left(\theta \right)$.

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

 

Simplify

$$\begin{gathered} e^{\left(zi\right)}-e^{\left(-zi\right)}=(2i-1)\left[e^{\left(zi\right)}+e^{\left(-zi\right)}\right] \\ e^{\left(2zi\right)}-1=(2i-1)e^{\left(2zi\right)}+(2i-1) \\ e^{\left(2zi\right)} = \frac{i}{1-i} \end{gathered}$$ 

Step 4: Convert in rectangular form.

Take the logarithm function both sides.

$$\begin{gathered} 2zi=\left(\frac{i}{1-i}\right) \\ 2z=\ln \left(\frac{exp\left(\mathrm{\pi i}/2\right)}{\sqrt{2}exp\left(-\mathrm{\pi i}/4\right)}\right) \\ 2z=\ln \left(\frac{exp\left(3\mathrm{\pi i}/4\right)}{\sqrt{2}}\right) \\ 2z=\ln \left(\frac{1}{\sqrt{2}}\right)+i\left(\frac{3\mathrm{\pi }}{4}\pm 2n\mathrm{\pi }\right) \\ 2z=-\frac{\ln \left(2\right)}{2}+i\left(\frac{3\mathrm{\pi }}{4}\pm 2n\mathrm{\pi }\right) \end{gathered}$$

where n = 0,1,2,3,....

 $$\begin{gathered} z=\frac{2zi}{2i} \\ z=\frac{-{\displaystyle \frac{\ln \left(2\right)}{2}}+i\left(\frac{3\mathrm{\pi }}{4}\pm 2n\mathrm{\pi }\right)}{2i} \\ z=\frac{i \ln \left(2\right)}{4}+\left(\frac{3\mathrm{\pi }}{8}\pm 2n\mathrm{\pi }\right) \end{gathered}$$

              

Hence the general solution equation arctan ( 2 + i ) is $\frac{i \ln \left(2\right)}{4}+\left(\frac{3\mathrm{\pi }}{8}\pm 2n\mathrm{\pi }\right)$.