Q24P Show that (ab)c聽 can have more ... [FREE SOLUTION]

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Chapter 2: Q24P (page 74)

Show that ${(a^{b})}^{c}$ can have more values than. ${a^{b}}^{c}$ As examples compare
${[{(鈭� i)}^{2 + i}]}^{2 鈭� i} and {(鈭� i)}^{(2 + i)}^{(2 鈭� i)} = {(鈭� i)}^{5}$ ;
${(i^{i})}^{i} and {(鈭� i)}^{鈭� 1}$ .

Short Answer

Expert verified

(a) Hence, ${[{(鈭� i)}^{2 + i}]}^{2 鈭� i}$ will have more value than. ${(鈭� i)}^{5}$

(b) Hence, ${(i^{i})}^{i}$ will have more value than. ${(i)}^{鈭� 1}$

Step by step solution

Complex Roots and Powers

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

Step 2:(a) Determine the proof

The given expressions are:. ${[{(鈭� i)}^{2 + i}]}^{2 鈭� i} and {(鈭� i)}^{(2 + i)}^{(2 鈭� i)} = {(鈭� i)}^{5}$

Letus take:

$\begin{array}{l} x_{1} = {[{(鈭� i)}^{2 + i}]}^{2 鈭� i} \\ x_{2} = {(鈭� i)}^{(2 + i)}^{(2 鈭� i)} = {(鈭� i)}^{5} \end{array}$

Evaluate $x_{2}$ as follow:

$\begin{matrix} x_{2} = {(鈭� i)}^{5} \\ = 鈭� i^{4} 鈰� i \\ = 鈭� i \end{matrix}$

For $x_{1}$ , let . $z = {(鈭� i)}^{2 + i}$ Then, using , $a^{b} = e^{b \ln a}$ we have

$\begin{matrix} z = {(鈭� i)}^{2 + i} \\ = e^{(2 + i) \ln (鈭� i)} \\ = e^{(2 + i) [\ln 1 + i (\frac{3 蟺}{2} 卤 2 n 蟺)]} 鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌� ......... {鈭赌 n 鈮� 0} \\ = e^{[i (3 蟺卤 4 n 蟺)]} 鈰� e^{[鈭� (\frac{3 蟺}{2} 卤 2 n 蟺)]} \\ = 鈭� e^{[鈭� (\frac{3 蟺}{2} 卤 2 n 蟺)]} 鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆€夆€夆€夆€夆€夆€夆€夆€夆€夆€� ........ {e^{[i (3 蟺卤 4 n 蟺)]} = 鈭� 1} \end{matrix}$

Now, we have:

$\begin{matrix} x_{1} = {[{(鈭� i)}^{2 + i}]}^{2 鈭� i} \\ = {[鈭� e^{[鈭� (\frac{3 蟺}{2} 卤 2 n 蟺)]}]}^{2 鈭� i} \end{matrix}$

Clearly, $x_{1}$ is greater than . $x_{2}$

Hence, ${[{(鈭� i)}^{2 + i}]}^{2 鈭� i}$ will have more value than . ${(鈭� i)}^{5}$

Step 3:(b)Determine the proof

The given expressions are: ${(i^{i})}^{i} and {(i)}^{鈭� 1}$ .

Let us take:

$\begin{array}{l} x_{1} = {(i^{i})}^{i} \\ x_{2} = {(i)}^{鈭� 1} \end{array}$

Evaluate $x_{2}$ as follow:

$\begin{matrix} x_{2} = {(i)}^{鈭� 1} \\ = \frac{1}{i} \\ = \frac{i}{i^{2}} \\ = 鈭� i \end{matrix}$

For, $x_{1}$ let . $z = {(i)}^{i}$ Then, using , $a^{b} = e^{b \ln a}$ we have

$\begin{matrix} z = i^{i} \\ = e^{i \ln (i)} \\ = e^{[i 鈰� i (\frac{蟺}{2} 卤 2 n 蟺)]} 鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌� ......... {鈭赌 n 鈮� 0} \\ = e^{[鈭� (\frac{蟺}{2} 卤 2 n 蟺)]} \end{matrix}$

Now, we have:

$\begin{matrix} x_{1} = {[{(i)}^{i}]}^{i} \\ = {[e^{[鈭� (\frac{蟺}{2} 卤 2 n 蟺)]}]}^{i} \end{matrix}$

Clearly, $x_{1}$ is greater than . $x_{2}$

Hence, ${(i^{i})}^{i}$ will have more value than. ${(i)}^{鈭� 1}$

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91影视

Show that ${(a^{b})}^{c}$ can have more values than. ${a^{b}}^{c}$ As examples compare
${[{(鈭� i)}^{2 + i}]}^{2 鈭� i} and {(鈭� i)}^{(2 + i)}^{(2 鈭� i)} = {(鈭� i)}^{5}$ ;
${(i^{i})}^{i} and {(鈭� i)}^{鈭� 1}$ .

Short Answer

Step by step solution

Complex Roots and Powers

Step 2:(a) Determine the proof

Step 3:(b)Determine the proof

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

Fundamentals of Physics

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91影视

Show that (ab)c can have more values than.abcAs examples compare[(鈭�i)2+i]2鈭�iand(鈭�i)(2+i)(2鈭�i)=(鈭�i)5;(ii)iand(鈭�i)鈭�1.

Short Answer

Step by step solution

Complex Roots and Powers

Step 2:(a) Determine the proof

Step 3:(b)Determine the proof

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

Fundamentals of Physics

Fluids

Physical Quantities And Units

Measurements

Geometrical and Physical Optics

Study anywhere. Anytime. Across all devices.

Show that ${(a^{b})}^{c}$ can have more values than. ${a^{b}}^{c}$ As examples compare
${[{(鈭� i)}^{2 + i}]}^{2 鈭� i} and {(鈭� i)}^{(2 + i)}^{(2 鈭� i)} = {(鈭� i)}^{5}$ ;
${(i^{i})}^{i} and {(鈭� i)}^{鈭� 1}$ .