Introduction to Complex Numbers and Algebra

The Imaginary Unit, Iota ($i$)

The real number system is insufficient to solve all algebraic equations, such as $x^2 + 1 = 0$. To address this, we define the imaginary unit, denoted by $i$ (iota), with the following properties:

$$i = \sqrt{-1} \quad \text{and} \quad i^2 = -1$$

What is a Complex Number?

A complex number, usually denoted by $z$, is any number that can be expressed in the form $a + ib$, where $a$ and $b$ are real numbers.

• Real Part: The real number $a$ is called the real part of $z$, denoted as Re($z$).
• Imaginary Part: The real number $b$ is called the imaginary part of $z$, denoted as Im($z$).

Equality of Complex Numbers

Two complex numbers, $z_1 = a+ib$ and $z_2 = c+id$, are equal if and only if their real parts are equal and their imaginary parts are equal.

$$a+ib = c+id \iff a=c \text{ and } b=d$$

Algebra of Complex Numbers

• Addition: $(a+ib) + (c+id) = (a+c) + i(b+d)$
• Subtraction: $(a+ib) - (c+id) = (a-c) + i(b-d)$
• Multiplication: $(a+ib)(c+id) = (ac-bd) + i(ad+bc)$
• Division: To divide, we multiply the numerator and denominator by the conjugate of the denominator: $\frac{a+ib}{c+id} = \frac{(a+ib)(c-id)}{(c+id)(c-id)} = \frac{ac+bd}{c^2+d^2} + i\frac{bc-ad}{c^2+d^2}$

Integral Powers of Iota ($i$)

The powers of $i$ are cyclical with a period of 4 ($i, -1, -i, 1$). To find $i^n$ for any integer $n$, we use the remainder when $n$ is divided by 4. $i^n = i^r$, where $r$ is the remainder.

• A key property is that the sum of four consecutive powers of $i$ is zero: $i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0$.

Example 1: Simplifying Powers of Iota

Question: Find the value of $i^{135}$.

Solution: To simplify $i^{135}$, we divide 135 by 4. $135 = 4 \times 33 + 3$. The remainder is 3. Therefore, $i^{135} = i^3 = -i$.

Example 2: Sum of Powers of Iota

Question: Evaluate the sum $i^{10} + i^{11} + i^{12} + i^{13}$.

Solution: This is a sum of four consecutive powers of iota. According to the property $i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0$, the sum is 0. Alternatively, we can calculate each term: $i^{10}=i^2=-1$, $i^{11}=i^3=-i$, $i^{12}=i^4=1$, $i^{13}=i^1=i$. The sum is $-1 - i + 1 + i = 0$.

Example 3: Multiplication of Complex Numbers

Question: Express the product $(1-i)(2+3i)$ in the form $a+ib$.

Solution:

$$(1-i)(2+3i) = 1(2) + 1(3i) - i(2) - i(3i) = 2 + 3i - 2i - 3i^2$$

Since $i^2 = -1$, this becomes:

$$= 2 + i - 3(-1) = 2 + i + 3 = 5 + i$$

Example 4: Equality of Complex Numbers

Question: Find the real values of x and y if $(x^4 + 2xi) - (3x^2 + iy) = (3 - 5i) + (1 + 2iy)$.

Solution: First, group the real and imaginary parts on both sides. LHS: $(x^4 - 3x^2) + i(2x - y)$ RHS: $(3+1) + i(-5 + 2y) = 4 + i(-5+2y)$ Equating real parts: $x^4 - 3x^2 = 4 \implies x^4 - 3x^2 - 4 = 0$. Let $u=x^2$, so $u^2-3u-4=0 \implies (u-4)(u+1)=0$. Thus $x^2=4$ or $x^2=-1$. Since x is real, $x^2=4 \implies x = \pm 2$. Equating imaginary parts: $2x - y = -5 + 2y \implies 2x + 5 = 3y$. If $x=2$, then $2(2)+5 = 3y \implies 9=3y \implies y=3$. If $x=-2$, then $2(-2)+5 = 3y \implies 1=3y \implies y=1/3$. The solutions are (x=2, y=3) and (x=-2, y=1/3).

Example 5: Finding a Multiplicative Inverse

Question: Find the multiplicative inverse of $z = 4 - 3i$.

Solution: The multiplicative inverse is $z^{-1} = 1/z$. We calculate this by multiplying the numerator and denominator by the conjugate of z.

$$z^{-1} = \frac{1}{4-3i} = \frac{1(4+3i)}{(4-3i)(4+3i)} = \frac{4+3i}{4^2 - (3i)^2} = \frac{4+3i}{16 - 9i^2} = \frac{4+3i}{16+9} = \frac{4+3i}{25}$$

So, the inverse is $\frac{4}{25} + i\frac{3}{25}$.

Example 6: Simplifying a Complex Fraction

Question: Express $\frac{5+i}{2-3i}$ in the form $a+ib$.

Solution: We multiply the numerator and denominator by the conjugate of the denominator, which is $2+3i$.

$$\frac{5+i}{2-3i} = \frac{(5+i)(2+3i)}{(2-3i)(2+3i)} = \frac{10+15i+2i+3i^2}{2^2+3^2} = \frac{10+17i-3}{13} = \frac{7+17i}{13}$$

The result is $\frac{7}{13} + i\frac{17}{13}$.

Example 7: Square Root of a Complex Number

Question: Find the square root of $7 + 24i$.

Solution: Let $\sqrt{7+24i} = x+iy$. Squaring both sides: $7+24i = (x+iy)^2 = (x^2-y^2) + i(2xy)$. Equating real and imaginary parts: 1) $x^2 - y^2 = 7$ 2) $2xy = 24 \implies xy = 12$ We also use the identity $(x^2+y^2)^2 = (x^2-y^2)^2 + (2xy)^2$. $(x^2+y^2)^2 = 7^2 + 24^2 = 49 + 576 = 625$. So, $x^2+y^2 = 25$ (since it must be positive). Now we have: $x^2+y^2=25$ $x^2-y^2=7$ Adding these two equations gives $2x^2=32 \implies x^2=16 \implies x = \pm 4$. Subtracting gives $2y^2=18 \implies y^2=9 \implies y = \pm 3$. Since $xy=12$ is positive, x and y must have the same sign. So the roots are $(x=4, y=3)$ and $(x=-4, y=-3)$. The square roots are $4+3i$ and $-4-3i$.