Understanding Complex Numbers: Real Part, Imaginary Unit, and Imaginary Part

Equation	Unsolvable in	Solvable in
$x + 2 = 1$	Natural numbers $N$	Integer numbers $Z$
$2 x - 3 = 0$	Integer numbers $Z$	Rational numbers $Q$
$x^{2} = 5$	Rational numbers $Q$	Irrational numbers
$x^{2} = - 2$	Real numbers $R$	$?$

Equation

Unsolvable in

Solvable in

x + 2 = 1

Natural numbers

N

Integer numbers

Z

2 x - 3 = 0

Integer numbers

Z

Rational numbers

Q

x^{2} = 5

Rational numbers

Q

Irrational numbers

x^{2} = - 2

Real numbers

R

?

	Square	Triangle
Area Formula	$A_{□} = s^{2}$	$A_{△} = \frac{1}{2} b h$
Substitution	$A_{□} = x^{2}$	$A_{△} = \frac{1}{2} \cdot 3 \cdot 4 = 6$

Square

Triangle

Area Formula

A_{□} = s^{2}

A_{△} = \frac{1}{2} b h

Substitution

A_{□} = x^{2}

A_{△} = \frac{1}{2} \cdot 3 \cdot 4 = 6

	Square	Triangle
Area	$A_{□} = x^{2}$	$A_{△} = 6$

Square

Triangle

Area

A_{□} = x^{2}

A_{△} = 6

Finding $i^{n}$
If the remainder of $\frac{n}{4}$ is	Then, $i^{n}$ is equal to
$0$	$i^{0} = 1$
$1$	$i^{1} = i$
$2$	$i^{2} = - 1$
$3$	$i^{3} = - i$

Finding

i^{n}

If the remainder of

\frac{n}{4}

Then,

i^{n}

is equal to

0

i^{0} = 1

1

i^{1} = i

2

i^{2} = - 1

3

i^{3} = - i

Component	Impedance
Resistor	$8 Ω$
Capacitor	$- 6 i Ω$
Inductor	$10 i Ω$

Component

Impedance

Resistor

8 Ω

Capacitor

- 6 i Ω

Inductor

10 i Ω

Component	Resistance or Reactance	Impedance
Resistor $1$	$6 Ω$	$6 Ω$
Capacitor	$9 Ω$	$- 9 i Ω$
Inductor	$4 Ω$	$4 i Ω$
Resistor $2$	$3 Ω$	$3 Ω$

Component

Resistance or Reactance

Impedance

Resistor

1

6 Ω

6 Ω

Capacitor

9 Ω

- 9 i Ω

Inductor

4 Ω

4 i Ω

Resistor

2

3 Ω

3 Ω

Pair the given powers of

i

with their simplified forms.

The Product of Powers Property holds true for the imaginary unit. We can use it to find the powers of i. Let's consider the first four powers of i. Recall that any number raised to the power of one equals itself, so i^1= i. By definition, i is equal to sqrt(- 1), so we know that i^2= - 1. l i^1= i i^2= - 1 i^3= i^2* i^1= - 1* i = - i i^4= i^2* i^2= - 1*( - 1)= 1 Note that the fourth power of i is equal to 1. Therefore, we can simplify the given powers of i by rewriting them so that i^4 is a factor. This will make the process in simplifying the given expressions easier! Using this trick together with the Product of Powers Property, let's calculate i^6.

We will compute i^(35) by applying the same procedure.

Let's now compute i^(1872).

Finally, we will simplify i^(1905).

Let's take a final look at all the pairings! i^6 &→ - 1 i^(35) &→ - i i^(1872) &→ 1 i^(1905) &→ i

Electrical circuit components such as resistors, inductors, and capacitors oppose the current flow. For resistors, this opposition is called resistance. For capacitors and inductors, it is called reactance. The total electrical impedance of a series circuit is equal to the sum of the opposition of each component.

Component	Resistance or Reactance $(Ω)$	Impedance $(Ω)$
Resistor	$R$	$R$
Inductor	$L$	$L i$
Capacitor	$C$	$- C i$

With this in mind, consider the following circuits.

What is the impedance of the circuit?

What is the impedance of the series circuit?

The impedance for a circuit is the sum of the impedances for the individual components. Impedance of Circuit: R+Li+(- Ci) In this case, the contribution of each component to the impedance of the circuit can be listed as below.

Component	Resistance or Reactance (Ω)	Impedance (Ω)
Resistor	12	12
Inductor	6	6i
Capacitor	7	- 7i

Now that we know the impedance of each component, we can find the total impedance of the circuit.

The impedance of the circuit is 12-i ohms.

In this case, we will again start by listing the contribution of each component to the impedance of the circuit.

Component	Resistance or Reactance (Ω)	Impedance (Ω)
Resistor I	14	14
Resistor II	4	4
Inductor	9	9i
Capacitor	2	- 2i

We will calculate the impedance of the circuit by adding up the impedances of the individual components.

The impedance of the circuit is 18+7i ohms.

Consider the following complex numbers.

z_{1} = 9 + 5 i z_{2} = 3 - 2 i

Calculate each operation and write the result in standard form.

z_{1} \cdot z_{2}

z_{2} \cdot \overset{z}{ˉ}_{2}

\frac{z _{1}}{z _{2}}

We will first substitute z_1=9+5i and z_2=3-2i into the given operation. Then we can multiply the expressions.

We will now simplify the expression by combining the real parts and the imaginary parts separately.

We will start by finding the conjugate of z_2. Recall that the conjugate of a complex number z, denoted by z, is found by changing the sign of the imaginary part of z. z_2=3-2i = 3+( -2)i The imaginary part of z_2 is - 2. Therefore, its conjugate will have 2 as its imaginary part. z_2=3+2i We can now substitute z_2= 3-2i and z_2= 3+2i into the given operation. z_2* z_2=( 3-2i)( 3+2i) Recall the fact that the product of a complex number and its conjugate is a real number. (a+bi)(a-bi) = a^2+b^2 We can use this fact to simplify the given product.

We will start by substituting z_1= 9+5i and z_2= 3-2i into z_1z_2. z_1/z_2=9+5i/3-2i We need to rationalize the denominator. To do so, we will multiply both numerator and denominator of the fraction by the conjugate of the denominator. We already found the conjugate of z_2 in Part B. z_2=3+2i We will now multiply the numerator and denominator of the given fraction by 3+2i. 9+5i/3-2i=(9+5i)( 3+2i)/(3-2i)( 3+2i) In Part B we also found that the product in the denominator is 13. This will help us in simplifying the quotient.

Now we need to write this expression in the standard form of a complex number. We can do this by rewriting it as a sum of fractions. 17+33i/13 ⇒ 17/13 + 33/13i

The voltage $V$ of a circuit is related to the electric current $C$ and impedance $I$ by the formula $V = C \cdot I .$ Use this formula to solve the following problems. Write the answers in standard form.

The current in the circuit is

2 + 4 i

amps and the impedance is

1 - 5 i

ohms. What is the voltage of the circuit?

The voltage in the circuit is

30 + 12 i

volts and the current is

5 - i

amps. What is the impedance of the circuit?

To find the voltage of the circuit, let's substitute the given current C=2+4i amps and the given impedance I=1-5i ohms into the formula V=C* I. We will then simplify the expression.

The voltage of the circuit is 22-6i volts.

To find the impedance of the circuit, let's substitute the given voltage V=30+12i volts and the given current C=5-i amps into the formula V=C* I. We will then solve for I.

To calculate the quotient, we need to rationalize the denominator. To do so, we will multiply both the numerator and the denominator of the fraction by the conjugate of the denominator. Denominator: & 5-i Conjugate: & 5+i We will now multiply the numerator and denominator of the fraction by 5+i. 30+12i/5-i=(30+12i)( 5+i)/(5-i)( 5+i) Using the fact that (a+bi)(a-bi)=a^2+b^2, we can rewrite the product in the denominator as 5^2+1^2, which results in a real number. Then we can simplify the quotient and write the result in standard form.

The impedance of the circuit is 6913 + 4513i ohms.

Emily plots the numbers

- 5,

1,

- 2 - 2 i,

and

- 2 + 5 i

on a complex plane and then connects them so that she gets a quadrilateral. What type of quadrilateral does Emily get?

We are given four complex numbers. - 5, 1, -2-2i, -2+5i Recall that in the complex plane, the horizontal axis represents the real part and the vertical axis the imaginary part of a complex number. This means that a complex number a+bi corresponds to the point (a,b). ccc Complex Number& &Point on Complex Plane - 5 & & (- 5, 0) 1 & & (1,0) -2-2i & & (-2,-2) -2+5i & & (-2,5) Let's plot these points on the complex plane.

Now let's connect the points so that we get a quadrilateral. To do so, we start at any of the points and connect them counterclockwise.

As we can see in the graph, the quadrilateral formed by these points is a kite.

The Basics of Complex Numbers

Catch-Up and Review

Non-real Solutions to Quadratic Equations

Expanding the Real Numbers Set

The Imaginary Unit

Rewriting Square Roots of Negative Numbers

Extra

Real World Situations With Imaginary Solutions

Hint

Solution

Powers of $i$

Looking for a Pattern

Calculating Powers of $i$

Imaginary and Complex Numbers

Imaginary Numbers

Complex Numbers

Combining Complex Numbers

Adding and Subtracting Complex Numbers

Impedance for a Series Circuit

Hint

Solution

Multiplication of Complex Numbers

Multiplying Complex Numbers

Voltage of Series Circuit

Hint

Solution

Dividing Complex Numbers

Complex Conjugate

Rationalizing a Denominator Using Complex Conjugates

Conjugates and Divisions

Hint

Solution

Operating With Complex Numbers

Extra

Graphing Complex Numbers in the Complex Plane

The Basics of Complex Numbers

Recommended exercises

	18 Theory slides
	14 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions