Q12 MP - Mathematical Methods in Physical Sciences

Question

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

${\mathit{e}}^{\mathbf{i}\mathbf{ }\mathbf{arcsin}\mathbf{ }\mathbf{i}}$

Step-by-Step Solution

Answer

The values of ${\mathit{e}}^{\mathbf{i}\mathbf{ }\mathbf{arcsin}\mathbf{ }\mathbf{i}}$are$0.414,-2.414$.

Step 1: Given Information

The given expression is $e^{i arc\sin i}$.       

Step 2: Definition of Complex Numbers

The numbers that are presented in the form of $\mathit{x}\mathbf{+}\mathit{i}\mathit{y}$where, $\mathbf{\text{'}}\mathit{x}\mathbf{\text{'}}$ is real numbers and $\mathbf{\text{'}}\mathit{i}\mathit{y}\mathbf{\text{'}}$ is an imaginary number, those numbers are referred to as called Complex numbers.

Step 3: Use the exponential form of tan ( z ) .

Consider $w=e^{i arc\sin i}$

 

Let $\arcsin \mathit{\left(}i\mathit{\right)}$be $z$.

           $z=\arcsin \left(i\right)$

  $\sin \left(z\right)=i$                       …(1)

                  $w=e^{\left(iz\right)}$            …(2)

 

The exponential form of $\sin \left(z\right)=\frac{e^{\left(zi\right)}-e^{\left(-zi\right)}}{2i}$                     …(3)

 

Put equation (3) in (1).

       $\frac{e^{\left(zi\right)}-e^{\left(-zi\right)}}{2i}=i$                                          …(4)

Step 4: Simplify the exponential form

Simplify equation (4).

$$\begin{gathered} e^{\left(zi\right)}-e^{\left(-zi\right)}=-2 \\ {e}^{\left(2zi\right)}+2e^{\left(zi\right)}-1=0 \end{gathered}$$       

                                                                         .....(5)

Let $e^{iz}$be $u$.

$u^2+2u-1=0$

Solve by using quadratic formula $u=\frac{-b\underline{+}\sqrt{b^2-4ac}}{2a}$.

Coefficients of the quadratic equation are as follows.

$$\begin{gathered} a=1 \\ b=2 \\ c=-1 \end{gathered}$$

 

Substitute the coefficients into the quadratic formula.

$$\begin{gathered} u=\frac{-2\underline{+}\sqrt{4-4\left(-1\right)}}{2} \\ =-1\underline{+}\sqrt{2} \end{gathered}$$

The two solutions $u_1$, $u_2$ are $\left(-1+\sqrt{2}\right),\left(-1-\sqrt{2}\right)$respectively.

Step 5: Find the value of i z 1 and i z 2 .

Find the value of $i{z}_1$.

$$\begin{gathered} i{z}_1=\mathrm{In}\left(u_{\mathit{1}}\right) \\ =\mathrm{In}\left[e^{\left(z_{1}i\right)}\right] \\ =\mathrm{In}\left(\sqrt{2}-1\right) \\ {\mathrm{iz}}_1=\mathrm{In}\left(\sqrt{2}-1\right)\underline{+}2n\pi i \end{gathered}$$                              ....(6)                               Find the value of $i{z}_2$.

$$\begin{gathered} i{z}_2=\mathrm{In}\left(u_2\right) \\ =\mathrm{In}\left[e^{\left(z_{2}i\right)}\right] \\ =\mathrm{In}\left(-\sqrt{2}-1\right) \\ i{z}_2=\mathrm{In}\left(\sqrt{2}+1\right)+i\left(\mathrm{\pi }\underline{+}2\mathrm{n\pi }\right) \end{gathered}$$                 ......(7)                                                                                       

Step 6: Find the value of w 1 and w 2

Find the value of $w_1$.

   $w_1=e^{i{z}_1}$                                    …(8)

Put equation (6) in (8).

$$\begin{gathered} w_1=e^{\left[\mathrm{In}\left(\sqrt{2}-1\right)\underline{+}2n\pi i\right]} \\ =e^{\mathrm{In}\left(\sqrt{2}-1\right)}.e^{2n\mathrm{\pi i}} \\ =\sqrt{2}-1 \\ =0.414 \end{gathered}$$

 

Find the value $w_2$

   $w_2=e^{i{z}_2}$                                          …(9)

Put equation (7) in (9).

$$\begin{gathered} w_2=e^{\left[In\left(\sqrt{2}+1\right)+\left(\mathrm{\pi }\underline{+}2\mathrm{n\pi }\right)i\right]} \\ =e^{In\left(\sqrt{2}-1\right)}.e^{\left(\mathrm{\pi }\underline{+}2\mathrm{n\pi }\right)i} \\ =\left[\sqrt{2}-1\right].\left[\cos \left(\mathrm{\pi }\underline{+}2\mathrm{n\pi }\right)+i \sin \left(\mathrm{\pi }\underline{+}2\mathrm{n\pi }\right)\right] \\ =-2.414 \end{gathered}$$

 

Hence the values of $e^{i arc\sin i}$are $0.414,-2.414$.