Q23P - Mathematical Methods in Physical Sciences

Question

Evaluate each of the following in x + iy form, and compare with a computer solution.$\mathbf{(}\mathbf{1}\mathbf{-}\sqrt{\mathbf{2}\mathbf{i}}{\mathbf{)}}^{\mathbf{i}}$Hint: Find $\sqrt{\mathbf{2}\mathbf{i}}$ first.

Step-by-Step Solution

Answer

The expression $(1-\sqrt{2i}{)}^i$ have two values that are $e^{\left[iln\right(2+i\left)\right]}$and $e^{(-0.147\pi \pm 2n\pi )}·[0.7+0.72i]$ .

Step 1: Given information

The given expression is $(1-\sqrt{2i}{)}^i.$

Step 2: Definition of Complex Number.

The numbers that are presented in the form of $\mathbf{a}\mathbf{+}\mathbf{ib}$ where, $\mathbf{ }\mathbf{a}\mathbf{,}\mathbf{b}$ are real numbers and$\mathbf{ }\mathbf{\text{'}}\mathbf{i}\mathbf{\text{'}}$is an imaginary number called complex numbers.

Step 3: Finding the roots

Consider,$\mathrm{w}=(1-\sqrt{2\mathrm{i}}{)}^{\mathrm{i}}.$    ……. (1)

 

Find the roots of $z=2i$ .

 

The modulus of the complex number is$r=2$

The argument of the complex number is $arc\tan \left(\frac{-\sqrt{2}}{0}\right)=\frac{\pi }{2}$

The numbers are imaginary. Find the roots.

$$\begin{gathered} z_k=r^{1/n}{e}^{\left[i\left(\frac{\theta +2\pi k}{2}\right)\right]} \\ {z}_k=2^{1/2}{e}^{\left[i\left(\frac{\pi /2+2\pi k}{2}\right)\right]} \end{gathered}$$

 

Find the roots at $k=1,2.$ 

$$\begin{gathered} {\mathrm{z}}_1=\sqrt{2}\exp \left[\mathrm{i}\left(\frac{\mathrm{\pi }}{4}\right)\right] \\ =1+\mathrm{i} ........\left(2\right) \\ {\mathrm{z}}_2=\sqrt{2}\exp \left[\mathrm{i}\left(\frac{5\mathrm{\pi }}{4}\right)\right] \\ =-1-\mathrm{i} ...........\left(3\right) \end{gathered}$$

Step 4: Use the roots to find the desired values

Put equation (2) in (1).

$$\begin{gathered} w_1={\left(1-1-i\right)}^i \\ ={\left(-i\right)}^i \\ =e^{\left[In{\left(-i\right)}^i\right]} \\ =e^{\left[i In\left(-i\right)\right]} \\ {w}_1=e^{\left[i In1+i\left(3\pi /2\pm 2n\pi \right)\right]} \\ =e^{\left[\left(3\pi /2\pm 2n\pi \right)\right]} \end{gathered}$$

Put equation (3) in (1).

$$\begin{gathered} w_2=\left(1+1+i\right) \\ ={\left(2+i\right)}^i \\ =e^{\left[In{\left(2+i\right)}^i\right]} \\ =e^{\left[i In \left(2+i\right)\right]} \\ {w}_2=e^{{}^{\left[i\left( In\sqrt{5}+i\left(0.147\pi \pm 2n\pi \right)\right)\right]}} \\ =e^{{}^{\left(i In\sqrt{5}\right) }}·e^{ \left(0.147\pi \pm 2n\pi \right)} \\ =e^{ \left(0.147\pi \pm 2n\pi \right)}·\left[0.7+0.72i\right] \end{gathered}$$

 Hence $(1-\sqrt{2i}{)}^i$ have two values that are$e^{\left[iln\right(2+i\left)\right]}$ and $e^{(-0.147\pi \pm 2n\pi )}·[0.7+0.72i]$ .