Q24P - Mathematical Methods in Physical Sciences

Question

Show that ${\mathbf{\left(}{\mathbf{a}}^{\mathbf{b}}\mathbf{\right)}}^{\mathbf{c}}$  can have more values than.${{\mathit{a}}^{\mathbf{b}}}^{\mathbf{c}}$ As examples compare 

1. ${\mathbf{\left[}{\mathbf{(}\mathbf{-}\mathbf{i}\mathbf{)}}^{\mathbf{2}\mathbf{+}\mathbf{i}}\mathbf{\right]}}^{\mathbf{2}\mathbf{-}\mathbf{i}}\mathbf{\text{ and }}{{\mathbf{(}\mathbf{-}\mathbf{i}\mathbf{)}}^{\mathbf{(}\mathbf{2}\mathbf{+}\mathbf{i}\mathbf{)}}}^{\mathbf{(}\mathbf{2}\mathbf{-}\mathbf{i}\mathbf{)}}\mathbf{=}{\mathbf{(}\mathbf{-}\mathbf{i}\mathbf{)}}^{\mathbf{5}}$ ;
2. ${\mathbf{\left(}{\mathbf{i}}^{\mathbf{i}}\mathbf{\right)}}^{\mathbf{i}}\mathbf{\text{ and }}{\mathbf{(}\mathbf{-}\mathbf{i}\mathbf{)}}^{\mathbf{-}\mathbf{1}}$ .

Step-by-Step Solution

Answer

(a) Hence,  ${\left[{(-i)}^{2+i}\right]}^{2-i}$will have more value than.${(-i)}^5$

(b) Hence,  ${\left(i^i\right)}^i$will have more value than.${\left(i\right)}^{-1}$

Step 1: Complex Roots and Powers

For any complex numbers, let say ,$a\text{ and }b$ the definition of the complex power induces a formula as: ${\mathit{a}}^{\mathbf{b}}\mathbf{=}{\mathit{e}}^{\mathbf{b}\mathbf{l}\mathbf{n}\mathbf{a}}$, where.$a\ne e$

Step 2:(a) Determine the proof

The given expressions are: .${\left[{(-i)}^{2+i}\right]}^{2-i}\text{ and }{{(-i)}^{(2+i)}}^{(2-i)}={(-i)}^5$

Letus take: 

 $\begin{array}{l}{x}_1={\left[{(-i)}^{2+i}\right]}^{2-i}\\ x_2={{(-i)}^{(2+i)}}^{(2-i)}={(-i)}^5\end{array}$

Evaluate  $x_2$ as follow:

 $\begin{array}{c}{x}_2={(-i)}^5\\ =-i^4\cdot i\\ =-i\end{array}$

For$x_1$ , let .$z={(-i)}^{2+i}$ Then, using ,$a^b=e^{b\ln a}$ we have

$\begin{array}{c}z={(-i)}^{2+i}\\ =e^{(2+i)\ln (-i)}\\ =e^{(2+i)[\ln 1+i(\frac{3\pi }{2}\pm 2n\pi )]}\mathrm{.........}\{\forall n\ge 0\}\\ =e^{\left[i(3\pi \pm 4n\pi )\right]}\cdot e^{[-(\frac{3\pi }{2}\pm 2n\pi )]}\\ =-e^{[-(\frac{3\pi }{2}\pm 2n\pi )]}\mathrm{........}\{e^{\left[i(3\pi \pm 4n\pi )\right]}=-1\}\end{array}$

Now, we have:

$\begin{array}{c}{x}_1={\left[{(-i)}^{2+i}\right]}^{2-i}\\ ={[-e^{[-(\frac{3\pi }{2}\pm 2n\pi )]}]}^{2-i}\end{array}$ 

Clearly, $x_1$  is greater than .$x_2$

Hence,  ${\left[{(-i)}^{2+i}\right]}^{2-i}$will have more value than .${(-i)}^5$

Step 3:(b)Determine the proof

The given expressions are: ${\left(i^i\right)}^i\text{ and }{\left(i\right)}^{-1}$ .

Let us take: 

$\begin{array}{l}{x}_1={\left(i^i\right)}^i\\ x_2={\left(i\right)}^{-1}\end{array}$ 

 Evaluate$x_2$   as follow:

$\begin{array}{c}{x}_2={\left(i\right)}^{-1}\\ =\frac{1}{i}\\ =\frac{i}{i^2}\\ =-i\end{array}$

For,$x_1$ let .$z={\left(i\right)}^i$ Then, using ,$a^b=e^{b\ln a}$ we have

$\begin{array}{c}z=i^i\\ =e^{i\ln \left(i\right)}\\ =e^{[i\cdot i(\frac{\pi }{2}\pm 2n\pi )]}\mathrm{.........}\{\forall n\ge 0\}\\ =e^{[-(\frac{\pi }{2}\pm 2n\pi )]}\end{array}$

Now, we have:

$\begin{array}{c}{x}_1={\left[{\left(i\right)}^i\right]}^i\\ ={\left[e^{[-(\frac{\pi }{2}\pm 2n\pi )]}\right]}^i\end{array}$ 

Clearly,$x_1$   is greater than .$x_2$

Hence,${\left(i^i\right)}^i$  will have more value than.${\left(i\right)}^{-1}$