 maths > complex-number

Exponent and Root of Complex Numbers

what you'll learn...

Overview

Exponent of Complex Numbers

»  Play with the forms of the complex number

→  component form $a+ib$$a + i b$

→  polar form $r\left(\mathrm{cos}\theta +i\mathrm{sin}\theta \right)$$r \left(\cos \theta + i \sin \theta\right)$

→  exponent form $r{e}^{i\theta }$$r {e}^{i \theta}$

»  Convert to the standard form of complex numbers $a+ib$$a + i b$

→  eg: ${\left(a+ib\right)}^{c+id}$${\left(a + i b\right)}^{c + i d}$: convert $\left(a+ib\right)$$\left(a + i b\right)$ to polar form $r{e}^{i\theta }$$r {e}^{i \theta}$

→  eg: ${r}^{id}$${r}^{i d}$ : convert $r$$r$ to ${e}^{\mathrm{ln}r}$${e}^{\ln r}$

Power of i

»  To calculate ${i}^{n}$${i}^{n}$

Remember ${i}^{4}=1$${i}^{4} = 1$ and ${i}^{2}=-1$${i}^{2} = - 1$. Quickly derive the following using this. No need to memorize

→  express $n$$n$ in the form $n=4p+2q+r$$n = 4 p + 2 q + r$, where $p,q,r\in ℕ$$p , q , r \in \mathbb{N}$ natural numbers.

→  ${i}^{n}={\left(-1\right)}^{q}×{i}^{r}$${i}^{n} = {\left(- 1\right)}^{q} \times {i}^{r}$

exponent

Given ${z}_{1}={a}_{1}+i{b}_{1}$${z}_{1} = {a}_{1} + i {b}_{1}$ and ${z}_{2}={a}_{2}+i{b}_{2}$${z}_{2} = {a}_{2} + i {b}_{2}$, Can the exponent ${z}_{1}^{{z}_{2}}$${z}_{1}^{{z}_{2}}$ be written in $a+ib$$a + i b$ form?
Yes. By converting to polar form `z_1 = r e^(i theta)

Given ${z}_{1}={a}_{1}+i{b}_{1}$${z}_{1} = {a}_{1} + i {b}_{1}$ and ${z}_{2}={a}_{2}+i{b}_{2}$${z}_{2} = {a}_{2} + i {b}_{2}$, exponent
${z}_{1}^{z}_2$${z}_{1}^{z} _ 2$
$\quad \quad = {\left({r}_{1} {e}^{i {\theta}_{1}}\right)}^{{a}_{2} + i {b}_{2}}$
$\quad = {r}_{1}^{a} _ 2$ $\quad \times {r}_{1}^{i {b}_{2}}$ $\quad \times {e}^{i {\theta}_{1} {a}_{2}}$ $\quad \times {e}^{i {\theta}_{1} i {b}_{2}}$

$\quad = {r}_{1}^{a} _ 2$
$\quad \times {e}^{i {\theta}_{1} {a}_{2}}$ $\quad \times {e}^{- {\theta}_{1} {b}_{2}}$

$\quad = {r}_{1}^{a} _ 2 {e}^{- {\theta}_{1} {b}_{2}}$ $\quad \times {e}^{i \left({b}_{2} \ln {r}_{1} + {\theta}_{1} {a}_{2}\right)}$

The result is in the polar form and can be converted to coordinate form.

Given ${z}_{1}={a}_{1}+i{b}_{1}$${z}_{1} = {a}_{1} + i {b}_{1}$ and ${z}_{2}={a}_{2}+i{b}_{2}$${z}_{2} = {a}_{2} + i {b}_{2}$, Can the root $\sqrt[{z}_{2}]{{z}_{1}}$$\sqrt[{z}_{2}]{{z}_{1}}$ be computed in $a+ib$$a + i b$ form?
Yes. By considering root as power of $\left(\frac{1}{{z}_{2}}\right)$$\left(\frac{1}{z} _ 2\right)$

Given ${z}_{1}={a}_{1}+i{b}_{1}$${z}_{1} = {a}_{1} + i {b}_{1}$ and ${z}_{2}={a}_{2}+i{b}_{2}$${z}_{2} = {a}_{2} + i {b}_{2}$, root
$\sqrt[{z}_{2}]{{z}_{1}}$$\sqrt[{z}_{2}]{{z}_{1}}$
$\quad \quad = {z}_{1}^{\frac{1}{z} _ 2}$
$\quad \quad = {z}_{1}^{\frac{\overline{{z}_{2}}}{| {z}_{2} {|}^{2}}}$
By following the rules of exponent of a complex number, the root can be solved.

What is ${\left(1+i\right)}^{\frac{3}{2}}$${\left(1 + i\right)}^{\frac{3}{2}}$?

The answer is '${2}^{\frac{3}{4}}\left(\mathrm{cos}\left(3\frac{\pi }{8}\right)+i\mathrm{sin}\left(3\frac{\pi }{8}\right)\right)$${2}^{\frac{3}{4}} \left(\cos \left(3 \frac{\pi}{8}\right) + i \sin \left(3 \frac{\pi}{8}\right)\right)$'

summary

Exponent and Roots of Complex number
•  For ${z}_{1}^{{z}_{2}}$${z}_{1}^{{z}_{2}}$, convert ${z}_{1}$${z}_{1}$ to polar form $r{e}^{i\theta }$$r {e}^{i \theta}$
•  For ${a}^{ib}$${a}^{i b}$, convert $a$$a$ to ${e}^{\mathrm{ln}a}$${e}^{\ln a}$ form
•  For ${z}_{1}^{\frac{1}{{z}_{2}}}$${z}_{1}^{\frac{1}{z} _ 2}$, convert $\frac{1}{{z}_{2}}$$\frac{1}{z} _ 2$ to a complex number in numerator $\frac{\overline{{z}_{2}}}{{|{z}_{2}|}^{2}}$$\frac{\overline{{z}_{2}}}{| {z}_{2} {|}^{2}}$

To find exponent and root of complex numbers, the rules of numerical expression is used to arrive at the coordinate form $a+ib$$a + i b$.

power of i

The value of $i$$i$ is '$\sqrt{-1}$$\sqrt{- 1}$'

The value of ${i}^{2}$${i}^{2}$ is '$-1$$- 1$'

It is defined that $i=\sqrt{-1}$$i = \sqrt{- 1}$. Substitute that in the following
${i}^{2}$${i}^{2}$
$\quad \quad = {\left(\sqrt{- 1}\right)}^{2}$
$\quad \quad = {\left(- {1}^{\frac{1}{2}}\right)}^{2}$
$\quad \quad = - {1}^{\frac{2}{2}}$
$\quad \quad = - 1$

The value of ${i}^{3}$${i}^{3}$ is '$-i$$- i$'

${i}^{3}$${i}^{3}$
$\quad \quad = {i}^{2 + 1}$
$\quad \quad = {i}^{2} \times {i}^{1}$
$\quad \quad = - 1 \times i$
$\quad \quad = - i$

The value of ${i}^{4}$${i}^{4}$ is '$1$$1$'

${i}^{4}$${i}^{4}$
$\quad \quad = {i}^{2 + 2}$
$\quad \quad = {\left(i\right)}^{2} \times {i}^{2}$
$\quad \quad = - 1 \times - 1$
$\quad \quad = 1$

The value of ${i}^{0}$${i}^{0}$ is '$1$$1$'

${i}^{0}$${i}^{0}$
$\quad \quad = {i}^{1 - 1}$
$\quad \quad = \frac{i}{i}$
$\quad \quad = 1$

The value of ${i}^{n}$${i}^{n}$, if $n=4p+2q+r$$n = 4 p + 2 q + r$ is '${\left(-1\right)}^{q}×{i}^{r}$${\left(- 1\right)}^{q} \times {i}^{r}$'

${i}^{n}$${i}^{n}$
$\quad \quad = {i}^{4 p + 2 q + r}$
$\quad \quad = {i}^{4 p} \times {i}^{2 q} \times {i}^{r}$

$\quad \quad = {\left(- 1\right)}^{q} \times {i}^{r}$

examples

What is the value of ${i}^{27}$${i}^{27}$?
The answer is '$-i$$- i$'

What is ${i}^{20}$${i}^{20}$?
The answer is '$1$$1$'

What is ${i}^{7}$${i}^{7}$?
The answer is '$-i$$- i$'

summary

Power of i: To calculate ${i}^{n}$${i}^{n}$, express $n$$n$ in the form $n=4p+2q+r$$n = 4 p + 2 q + r$, where $p,q,r\in ℕ$$p , q , r \in \mathbb{N}$ natural numbers. Then
${i}^{n}={\left(-1\right)}^{q}×{i}^{r}$${i}^{n} = {\left(- 1\right)}^{q} \times {i}^{r}$.

To calculate ${i}^{n}$${i}^{n}$, use ${i}^{4}=1$${i}^{4} = 1$ and ${i}^{2}=-1$${i}^{2} = - 1$.

Outline