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. ${\mathbf{\left(}{\mathbf{i}}^{\mathbf{i}}\mathbf{\right)}}^{\mathbf{i}}\mathbf{\text{and}}{\mathbf{(}\mathbf{-}\mathbf{i}\mathbf{)}}^{\mathbf{-}\mathbf{1}}$.

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.$a\ne e$

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 .${(-i)}^5$

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.${\left(i\right)}^{-1}$