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Algebraic Operations

Arithmetic in the Complex Plane

Working with complex numbers, of the form a+bia + bi, feels a lot like working with familiar algebraic expressions. The operations of addition, subtraction, multiplication, and division are all well-defined and follow consistent rules. They also have a beautiful geometric interpretation on the complex plane, which can make them easier to visualize.

Let's start with addition and subtraction. To add or subtract two complex numbers, you simply combine their corresponding real and imaginary parts. Think of it just like combining like terms in algebra.

(a+bi)+(c+di)=(a+c)+(b+d)i(a+bi)(c+di)=(ac)+(bd)i\begin{aligned} \\ (a+bi) + (c+di) &= (a+c) + (b+d)i \\ (a+bi) - (c+di) &= (a-c) + (b-d)i \\ \end{aligned}

For example, to add 3+2i3 + 2i and 1+4i1 + 4i, we get (3+1)+(2+4)i=4+6i(3+1) + (2+4)i = 4 + 6i. Geometrically, this is identical to adding vectors. If you draw an arrow from the origin to 3+2i3 + 2i and another to 1+4i1 + 4i, their sum is the diagonal of the parallelogram they form.

Multiplication and Division

Multiplication also uses a familiar algebraic tool: the distributive property. You multiply the terms just as you would with binomials, remembering that i2=1i^2 = -1. This single rule is the key to the whole process.

(a+bi)(c+di)=a(c+di)+bi(c+di)=ac+adi+bci+bdi2=ac+(ad+bc)i+bd(1)=(acbd)+(ad+bc)i\begin{aligned} \\ (a+bi)(c+di) &= a(c+di) + bi(c+di) \\ &= ac + adi + bci + bdi^2 \\ &= ac + (ad+bc)i + bd(-1) \\ &= (ac-bd) + (ad+bc)i \\ \end{aligned}

For example: (2+3i)(4i)=2(4)+2(i)+3i(4)+3i(i)=82i+12i3i2=8+10i3(1)=11+10i(2+3i)(4-i) = 2(4) + 2(-i) + 3i(4) + 3i(-i) = 8 - 2i + 12i - 3i^2 = 8 + 10i - 3(-1) = 11+10i.

Division introduces a new, essential concept: the s. The conjugate of a complex number z=a+biz = a + bi is denoted as zˉ\bar{z} and is defined as zˉ=abi\bar{z} = a - bi. Essentially, you just flip the sign of the imaginary part.

Geometrically, the conjugate is a reflection across the real axis. The key property of a conjugate is what happens when you multiply a complex number by it: the result is always a real number.

zzˉ=(a+bi)(abi)=a2(bi)2=a2b2i2=a2+b2z\bar{z} = (a+bi)(a-bi) = a^2 - (bi)^2 = a^2 - b^2i^2 = a^2 + b^2

We use this trick to divide complex numbers. To compute a+bic+di\frac{a+bi}{c+di}, we want to make the denominator real. We can achieve this by multiplying both the numerator and the denominator by the conjugate of the denominator. This process is similar to rationalizing the denominator when working with square roots.

To divide two complex numbers, we multiply the numerator and denominator by the conjugate of the denominator to eliminate the imaginary part from the denominator.

a+bic+di=a+bic+dicdicdi=(ac+bd)+(bcad)ic2+d2\frac{a+bi}{c+di} = \frac{a+bi}{c+di} \cdot \frac{c-di}{c-di} = \frac{(ac+bd) + (bc-ad)i}{c^2+d^2}

Powers of i

Finally, let's look at the powers of the imaginary unit, ii. They follow a simple, repeating cycle of four values. This pattern makes calculating high powers of ii much easier than you might expect.

PowerCalculationResult
i1i^1iiii
i2i^2iii \cdot i1-1
i3i^3i2ii^2 \cdot ii-i
i4i^4i2i2i^2 \cdot i^211
i5i^5i4ii^4 \cdot iii

As you can see, the pattern i,1,i,1i, -1, -i, 1 repeats every four powers. To find the value of ini^n for any integer nn, you can use the remainder when nn is divided by 4. For instance, to find i35i^{35}, we divide 35 by 4. The remainder is 3, so i35=i3=ii^{35} = i^3 = -i.

Let's test your understanding of these operations.

Quiz Questions 1/6

What is the sum of (5+2i)(5 + 2i) and (3+7i)(-3 + 7i)?

Quiz Questions 2/6

Calculate the product: (32i)(4+i)(3 - 2i)(4 + i)

These algebraic rules provide the foundation for working with complex numbers in any context, from solving equations to engineering applications.