Q11P - Mathematical Methods in Physical Sciences

Question

Find each of the following in the $\mathit{x}\mathbf{+}\mathit{i}\mathit{y}$ form and compare a computer solution.

$\mathit{s}\mathit{i}\mathit{n}{\mathit{h}}^{\mathbf{-}\mathbf{1}}\left(i I\sqrt{2}\right)$.

Step-by-Step Solution

Answer

The form of the given equation $\sin h^{-1 }\left(i I\sqrt{2}\right)$

$$\begin{gathered} z_1=i\left(\frac{\pi }{4}+2n\pi \right) \\ {z}_2=i\left(\frac{3\pi }{4}+2n\pi \right) \end{gathered}$$

Step 1: Given Information.

The given expression is,$\sin h^{-1}\left(i I\sqrt{2}\right)$ .

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 quadratic equation.

Consider the complex number $z=\sin h^{-1}\left(i I\sqrt{2}\right)$.

Rewrite the above expression.

$\sin h\left(z\right)=\frac{1}{\sqrt{2}}$

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

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

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

Put $e^{\left(zi\right)}=u$

$$\begin{gathered} u-\frac{1}{u}=i\sqrt{2} \\ {u}^2-i\sqrt{2u}-1=0 \end{gathered}$$

 

Step 4: Solve the quadratic equation.

Write the coefficient and then substitute in the formula.

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

 

Put in the formula.

$$\begin{gathered} u=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ u=\frac{i\sqrt{2}\pm \sqrt{-2+4}}{2} \\ {u}_1=\frac{i\sqrt{2}+\sqrt{2}}{2} \\ {u}_2=\frac{i\sqrt{2}-\sqrt{2}}{2} \end{gathered}$$

Step 5: Convert in rectangular form.

Convert in rectangular form.

 

Find the value of  $z_1$.

$z_1=In\left(u_1\right)$

Take $n=0,1,2,3,...$. for the values below.

$$\begin{gathered} z_1=In\left|\frac{i\sqrt{2}+\sqrt{2}}{2}\right|+i\left(\theta +2n\pi \right) \\ {z}_1=In\left(1\right)+i\left(\frac{\pi }{4}+2n\pi \right) \\ {z}_1=i\left(\frac{\pi }{4}+2n\pi \right) \end{gathered}$$

Step 6: Convert in rectangular form.

Find the value of $z_2$.

$$\begin{gathered} z^2=In\left(u_2\right) \\ {z}^2=In\left(r\right)+i\left(\theta +2n\pi \right) \\ {z}_2=In\left|\frac{i\sqrt{2}-\sqrt{2}}{2}\right|+i\left(\theta +2n\pi \right) \\ {z}_2=In\left(1\right)+i\left(\frac{3\pi }{4}+2n\pi \right) \\ {z}^2=i\left(\frac{3\pi }{4}+2n\pi \right) \end{gathered}$$

Hence the general solution of the given equation $\sin h^{-1}\left(i I\sqrt{2}\right)$

$$\begin{gathered} z_1=i\left(\frac{\pi }{4}+2n\pi \right) \\ {z}^2=i\left(\frac{3\pi }{4}+2n\pi \right) \end{gathered}$$