## 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