Complex Numbers : Exponent and Root of Complex Numbers

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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 $a + i b$ $a+ib$

    → polar form $r (\cos θ + i \sin θ)$ $r(cos theta + i sin theta)$

    → exponent form $r e^{i θ}$ $re^(i theta)$

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

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

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

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 $n = 4 p + 2 q + r$ $n=4p+2q+r$ , where $p, q, r \in ℕ$ $p,q, r in NN$ natural numbers.

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

exponent

Given $z_{1} = a_{1} + i b_{1}$ $z_1 = a_1+ib_1$ and $z_{2} = a_{2} + i b_{2}$ $z_2 = a_2+ib_2$ , Can the exponent $z_{1}^{z_{2}}$ $z_1^(z_2)$ be written in $a + i b$ $a+ib$ form?
Yes. By converting to polar form `z_1 = r e^(i theta)

Given $z_{1} = a_{1} + i b_{1}$ $z_1 = a_1+ib_1$ and $z_{2} = a_{2} + i b_{2}$ $z_2 = a_2+ib_2$ , exponent
$z_{1}^{z}_2$ $z_1^z_2$
$= {(r_{1} e^{i θ_{1}})}^{a_{2} + i b_{2}}$ $quad quad = (r_1e^(i theta_1) )^(a_2+ib_2)$
$= r_{1}^{a}_2$ $quad = r_1^a_2$ $\times r_{1}^{i b_{2}}$ $quad xx r_1^(ib_2)$ $\times e^{i θ_{1} a_{2}}$ $quad xx e^(i theta_1 a_2)$ $\times e^{i θ_{1} i b_{2}}$ $quad xx e^(i theta_1 i b_2)$

$= r_{1}^{a}_2$ $quad = r_1^a_2$
$\times e^{i b_{2} {\ln r}_{1}} ($ $quad xx e^(ib_2 ln r_1) ($ $\times e^{i θ_{1} a_{2}}$ $quad xx e^(i theta_1 a_2)$ $\times e^{- θ_{1} b_{2}}$ $quad xx e^(- theta_1 b_2)$

$= r_{1}^{a}_2 e^{- θ_{1} b_{2}}$ $quad = r_1^a_2 e^(- theta_1 b_2)$ $\times e^{i (b_{2} {\ln r}_{1} + θ_{1} a_{2})}$ $quad xx e^(i(b_2 ln r_1 + theta_1 a_2))$

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

Given $z_{1} = a_{1} + i b_{1}$ $z_1 = a_1+ib_1$ and $z_{2} = a_{2} + i b_{2}$ $z_2 = a_2+ib_2$ , Can the root $\sqrt[z_{2}]{z_{1}}$ $root(z_2)(z_1)$ be computed in $a + i b$ $a+ib$ form?
Yes. By considering root as power of $(\frac{1}{z_{2}})$ $(1/z_2)$

Given $z_{1} = a_{1} + i b_{1}$ $z_1 = a_1+ib_1$ and $z_{2} = a_{2} + i b_{2}$ $z_2 = a_2+ib_2$ , root
$\sqrt[z_{2}]{z_{1}}$ $root(z_2)(z_1)$
$= z_{1}^{\frac{1}{z_{2}}}$ $quad quad = z_1^(1/z_2)$
$= z_{1}^{\frac{\bar{z_{2}}}{{| z_{2} |}^{2}}}$ $quad quad = z_1^(bar(z_2)/(|z_2|^2))$
By following the rules of exponent of a complex number, the root can be solved.

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

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

summary

Exponent and Roots of Complex number
• For $z_{1}^{z_{2}}$ $z_1^(z_2)$ , convert $z_{1}$ $z_1$ to polar form $r e^{i θ}$ $re^(i theta)$
• For $a^{i b}$ $a^(ib)$ , convert $a$ $a$ to $e^{\ln a}$ $e^(ln a)$ form
• For $z_{1}^{\frac{1}{z_{2}}}$ $z_1^(1/z_2)$ , convert $\frac{1}{z_{2}}$ $1/z_2$ to a complex number in numerator $\frac{\bar{z_{2}}}{{| z_{2} |}^{2}}$ $bar(z_2)/(|z_2|^2)$

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

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
$i^{2}$ $i^2$
$= {(\sqrt{- 1})}^{2}$ $quad quad = (sqrt(-1))^2$
$= {(- 1^{\frac{1}{2}})}^{2}$ $quad quad = (-1^(1/2))^2$
$= - 1^{\frac{2}{2}}$ $quad quad = -1^(2/2)$
$= - 1$ $quad quad = -1$

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

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

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

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

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

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

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

$i^{n}$ $i^n$
$= i^{4 p + 2 q + r}$ $quad quad = i^(4p+2q+r)$
$= i^{4 p} \times i^{2 q} \times i^{r}$ $quad quad = i^(4p) xx i^(2q) xx i^r$
$= {(i^{4})}^{p}) \times {(i^{2})}^{q}) \times i^{r}$ $quad quad = (i^4)^p) xx (i^2)^q) xx i^r$
$= {(- 1)}^{q} \times i^{r}$ $quad quad = (-1)^q xx i^r$

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 $n = 4 p + 2 q + r$ $n=4p+2q+r$ , where $p, q, r \in ℕ$ $p,q, r in NN$ natural numbers. Then
$i^{n} = {(- 1)}^{q} \times i^{r}$ $i^n = (-1)^q xx i^r$ .

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

Outline

The outline of material to learn "complex numbers" is as follows.

Note : Click here for detailed overview of Complex-Numbers

→ Complex Numbers in Number System

→ Representation of Complex Number (incomplete)

→ Euler's Formula

→ Generic Form of Complex Numbers

→ Argand Plane & Polar form

→ Complex Number Arithmetic Applications

→ Understanding Complex Artithmetics

→ Addition & Subtraction

→ Multiplication, Conjugate, & Division

→ Exponents & Roots

→ Properties of Addition

→ Properties of Multiplication

→ Properties of Conjugate

→ Algebraic Identities