Q12P - Mathematical Methods in Physical Sciences

Question

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

${\mathbf{cosh}}^{\mathbf{-}\mathbf{1}}\left(\sqrt{3}/2\right)$

Step-by-Step Solution

Answer

The form of the given equation ${\cosh }^{-1}(\sqrt{3}/2)$
$$\begin{gathered} z_1=i\left(\frac{\mathrm{\pi }}{6}\pm 2n\mathrm{\pi }\right) \\ {z}_2=i\left(-\frac{\mathrm{\pi }}{6}\pm 2n\mathrm{\pi }\right) \end{gathered}$$

Step 1: Given Information.

The given expression is ${\cosh }^{-1}\left(\sqrt{3}/2\right)$.

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 is the real part and y is the imaginary part.

Step 3: Convert in quadratic equation.

Consider the complex number $\mathrm{z}={\cosh }^{-1}\left(\sqrt{3}/2\right)$.

Rewrite the above expression.

$\cosh \left(\mathrm{z}\right)=\frac{\sqrt{3}}{2}$

 

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

$\frac{e^{\left(z\right)}+e^{(-z)}}{2}=\frac{\sqrt{3}}{2}$

 

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

$$\begin{gathered} u+\frac{1}{u}=\sqrt{3} \\ {u}^2=-\sqrt{3}u+1=0 \end{gathered}$$ 

Step 4: Solve the quadratic equation.

Write the coefficient and then substitute in the formula.

$$\begin{gathered} a=1 \\ b=-\sqrt{3} \\ c=1 \end{gathered}$$

 

Put in the formula.

$$\begin{gathered} u=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ u=\frac{\sqrt{3}\pm \sqrt{3-4}}{2} \\ {u}_1=\frac{\sqrt{3}+i}{2} \\ {u}_2=\frac{\sqrt{3}-i}{2} \end{gathered}$$

Step 5: Convert in rectangular form.

Convert in rectangular form.

 

Find the value of $z_1$.

${\mathrm{z}}_1=\ln \left({\mathrm{u}}_1\right)$

 

Take n=0,1,23,.... for the values below.

$$\begin{gathered} z_1=\ln \left(\frac{\sqrt{3}+i}{2}\right) \\ {z}_1=\ln \left(1\right)+i\left(-\frac{\mathrm{\pi }}{6}\pm 2n\mathrm{\pi }\right) \\ {z}_1=i\left(\frac{\mathrm{\pi }}{6}\pm 2n\mathrm{\pi }\right) \end{gathered}$$

Step 6: Convert in rectangular form.

Find the value of $z_2$.

$$\begin{gathered} z_2=\ln \left(u_2\right) \\ {z}_2=\ln \left(\frac{\sqrt{3}-i}{2}\right) \\ {z}_2=\ln \left(1\right)+i\left(-\frac{\mathrm{\pi }}{6}\pm 2n\mathrm{\pi }\right) \\ {z}_2=i\left(-\frac{\mathrm{\pi }}{6}\pm 2n\mathrm{\pi }\right) \end{gathered}$$

 

Hence the general solution of the equation ${\cosh }^{-1}\left(\sqrt{3}/2\right)$

$$\begin{gathered} z_1=i\left(\frac{\mathrm{\pi }}{6}\pm 2n\mathrm{\pi }\right) \\ {z}_2=i\left(-\frac{\mathrm{\pi }}{6}\pm 2n\mathrm{\pi }\right) \end{gathered}$$