An Interactive Introduction to Complex Numbers

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3. Exponentiations

Example 3.1:

Open the Applet Exponentiations and set \(z = \sqrt 2 \cdot e^{22.5^\circ \cdot i}\) in exponential form. Then click on \(z^n\) and set \(n=2\). Look at the result in exponential form. Note that \(\left| {z^2 } \right| = \left| z \right|^2 = r^2\) and \(\arg \left( {z^2 } \right) = 2 \cdot \varphi\).
Indeed, we calculate \(z^2 = \left( {\sqrt 2 \cdot e^{22.5^\circ \cdot i} } \right)^2 = \left( {\sqrt 2 } \right)^2 \cdot \left( {e^{22.5^\circ \cdot i} } \right)^2 = 2 \cdot e^{22.5^\circ \cdot i \cdot 2} = 2 \cdot e^{45^\circ \cdot i}\).

Shift \(n\) to \(n=4\), which leads to \(z^4 = \left( {\sqrt 2 } \right)^4 \cdot \left( {e^{22.5^\circ \cdot i} } \right)^4 = 4 \cdot e^{22.5^\circ \cdot i \cdot 4} = 4 \cdot e^{90^\circ \cdot i}\).

Rule 3.1

The \(n\)-th exponentiation of \(z = r \cdot e^{\varphi \cdot i}\) in exponential form is \(z^n = \left( {r \cdot e^{\varphi \cdot i} } \right)^n = r^n \cdot \left( {e^{\varphi \cdot i} } \right)^n\) \(= r^n \cdot e^{n \cdot \varphi \cdot i} \quad n \in \mathbb{R}\) with \(\left| {z^n } \right| = \left| z \right|^n = r^n\) and \(\arg \left( {z^n } \right) = n \cdot \varphi\).

There exists no general rule for the exponentiation of \(z = a + b \cdot i\) in cartesian form. If \(n \in \mathbb{N}\), we could use the general binomial theorem. However, in most cases it is easier to transform \(z = a + b \cdot i\) into exponential form and apply rule 3.1.

In case of \(n=2\) the first binomial equation tells us that \(z^2 = \left( {a + b \cdot i} \right)^2 = a^2 + 2 \cdot a \cdot b \cdot i + \left( {b \cdot i} \right)^2\) \(= a^2 - b^2 + 2 \cdot a \cdot b \cdot i\).

Recall, we defined the exponent as \(n \in \mathbb{R}\), i. e. negative values of \(n\) are feasible as well.

Example 3.2:

Open the Applet Exponentiations and set \(z = \sqrt 2 \cdot e^{22.5^\circ \cdot i}\) in exponential form. Then click on \(z^n\) and set \(n=-2\). Of course, we get \(z^{ - 2} = \left( {\sqrt 2 \cdot e^{22.5^\circ \cdot i} } \right)^{ - 2} = \left( {\sqrt 2 } \right)^{ - 2} \cdot \left( {e^{22.5^\circ \cdot i} } \right)^{ - 2}\) \(= 0.5 \cdot e^{22.5^\circ \cdot i \cdot \left( { - 2} \right)} = 0.5 \cdot e^{ - 45^\circ \cdot i} = 0.5 \cdot e^{315^\circ \cdot i}\).

The result can also be derived by calculating \(z^{-2}= \frac{{1}}{{z^2}}\) just as with real numbers. Regarding example 3.1 we calculate \(z^{ - 2} = \frac{1}{{z^2 }} = \frac{1}{{2 \cdot e^{45^\circ \cdot i} }} = \frac{{1 \cdot e^{0^\circ \cdot i} }}{{2 \cdot e^{45^\circ \cdot i} }}\) \(= 0.5 \cdot e^{ - 45^\circ \cdot i}\) \(= 0.5 \cdot e^{315^\circ \cdot i}\).

Rule 3.2

\(z^{ - n} = \frac{1}{{z^n }}\quad n \in \mathbb{R},\;{\mathop{\rm Re}\nolimits} \left( z \right) \neq 0 \vee {\mathop{\rm Im}\nolimits} \left( z \right) \ne 0,\;\left| z \right| \ne 0\).


Roots

The natural question to ask now is, whether \(z = r \cdot e^{\varphi \cdot i}\) has also roots i.e. are there any complex numbers \(w\) solving for \(w^k = z\quad k \in \mathbb{N}\)?

Example 3.3 a:

From example 3.1 we can conclude that \( w = \sqrt 2 \cdot e^{22.5^\circ \cdot i} = 1.4142 \cdot e^{22.5^\circ \cdot i}\) is a square root of \(z = 2 \cdot e^{45^\circ \cdot i}\) or, alternatively, a solution of \(w^2 = 2 \cdot e^{45^\circ \cdot i}\). But is it the only one? To check this out, open the Applet Exponentiations and set \(z = 2 \cdot e^{45^\circ \cdot i}\) in exponential form. Then click on \(w^k=z\) and set \(k=2\). We see that \(w^2=2 \cdot e^{45^\circ \cdot i}\) has two solutions: \(w_1 = 1.4142 \cdot e^{22.5^\circ \cdot i}\) and \(w_2 = 1.4142 \cdot e^{202.5^\circ \cdot i}\).

What is the intuition behind the second solution \(w_2 = \sqrt 2 \cdot e^{202.5^\circ \cdot i}\)? Recall that \(z = 2 \cdot e^{45^\circ \cdot i}\) corresponds to \(z = 2 \cdot e^{405^\circ \cdot i}\). Then \(w_2 = \left( {2 \cdot e^{405^\circ \cdot i} } \right)^{0.5}\) \(= \sqrt 2 \cdot e^{405^\circ \cdot i \cdot 0.5} = 1.4142 \cdot e^{202.5^\circ \cdot i}\).

\(z = 2 \cdot e^{45^\circ \cdot i}\) also corresponds to \(z = 2 \cdot e^{765^\circ \cdot i}\), \(z = 2 \cdot e^{1'125^\circ \cdot i}\) and so on. But why do further solutions not exist? We see that \(w = \left( {2 \cdot e^{765^\circ \cdot i} } \right)^{0.5}\) \(= 1.4142 \cdot e^{382.5^\circ \cdot i}\), which corresponds to \(w_1 = \sqrt 2 \cdot e^{22.5^\circ \cdot i}\), whereas \(w = \left( {2 \cdot e^{1'125^\circ \cdot i} } \right)^{0.5}\) \(= 1.4142 \cdot e^{562.5^\circ \cdot i}\) corresponds to \(w_2 = 1.4142 \cdot e^{202.5^\circ \cdot i}\).

Example 3.3 b:

Open the Applet Exponentiations and set \(z = 2 \cdot e^{45^\circ \cdot i}\) in exponential form. Then click on \(w^k=z\) and set \(k=3\) and \(k=4\).
For \(k=3\) we get \(w_1 = 1.2599 \cdot e^{15^\circ \cdot i}\), \(w_2 = 1.2599 \cdot e^{135^\circ \cdot i}\) and \(w_3 = 1.2599 \cdot e^{255^\circ \cdot i}\).
For \(k=4\) we get \(w_1 = 1.1892 \cdot e^{11.25^\circ \cdot i}\), \(w_2 = 1.1892 \cdot e^{101.25^\circ \cdot i}\), \(w_3 = 1.1892 \cdot e^{191.25^\circ \cdot i}\) and \(w_4 = 1.1892 \cdot e^{281.25^\circ \cdot i}\).

Rule 3.3 a

\(w^k=z\) has \(k \in \mathbb{N}\) solutions. The angle between succeeding solutions is \(\frac{{360^\circ}}{{k}}\) (degree) or \(\frac{{2 \cdot \pi}}{{k}}\) (radian).


Rule 3.3 b

The first solution of \(w^k = z = r \cdot e^{\varphi \cdot i}\) is \(w_1 = \sqrt[k]{r} \cdot e^{\frac{\varphi }{k} \cdot i}\).


Rule 3.3 c

In degree the \(j\)-th solution of \(w^k = z = r \cdot e^{\varphi \cdot i}\) is \(w_j = \sqrt[k]{r} \cdot e^{\frac{{\varphi + \left( {j - 1} \right) \cdot 360^\circ }}{k} \cdot i}\) with \(\left| {w_j } \right| = \sqrt[k]{r}\) and \(\varphi _{wj} = \arg w_j = \frac{{\varphi + \left( {j - 1} \right) \cdot 360^\circ }}{k}\).
In radian the \(j\)-th solution of \(w^k = z = r \cdot e^{\varphi \cdot i}\) is \(w_j = \sqrt[k]{r} \cdot e^{\frac{{\varphi + \left( {j - 1} \right) \cdot 2 \cdot \pi }}{k} \cdot i}\) with \(\left| {w_j } \right| = \sqrt[k]{r}\) and \(\varphi _{wj} = \arg w_j = \frac{{\varphi + \left( {j - 1} \right) \cdot 2 \cdot \pi }}{k}\).

There is no general rule for roots of complex numbers in cartesian form.


Exercise 3.1

We have \(z = 1.5 \cdot e^{117^\circ \cdot i}\) and \(z= 5 - 2\cdot i\). Determine \(z^2\), \(z^{-3}\) and \(z^{0.6}\) for both numbers. Use the Applet Exponentiations to check your answers.

Exercise 3.2

Turn \(z=5\cdot i\) into exponential form. Then determine all solutions of \(w^5=z\) and rewrite them in cartesian form. Use the Applet Exponentiations to check your answers.

Dr. Jens Siebel, Last update: 08/15/2010