Complex Numbers

A2.C.1Imaginary unit & powers01/04

Imaginary unit and powers of i.

Step01of 04

The imaginary unit $i$ is defined to satisfy:

i^{2} = - 1

Equivalently, $i = - 1$ . This single definition extends real numbers to complex numbers and lets every quadratic equation have a solution.

A2.C.1 — Imaginary unit and powers of i.

Step 01 of 04

The imaginary unit $i$ is defined to satisfy:

i^{2} = - 1

Equivalently, $i = - 1$ . This single definition extends real numbers to complex numbers and lets every quadratic equation have a solution.

Step 02 of 04

Powers of i cycle every 4.

$i^{1}$	→ $i$
$i^{2}$	→ $- 1$
$i^{3}$	→ $- i$
$i^{4}$	→ $1$
$i^{5}$	→ $i$ (back to start)

For ANY power, find the remainder when dividing by 4. That tells you which of the four values you land on.

Step 03 of 04

Worked examples.

i^{27} : 27 \div 4 = 6 r 3 \Rightarrow i^{27} = i^{3} = - i

i^{50} : 50 \div 4 = 12 r 2 \Rightarrow i^{50} = i^{2} = - 1

i^{100} : 100 \div 4 = 25 r 0 \Rightarrow i^{100} = i^{0} = 1

Step 04 of 04

Negative powers. $i^{- 1} = 1/ i = - i$ (multiply by $i / i$ and use $i^{2} = - 1$ ). The same cycle applies but shifted.

Square roots of negative numbers. $- 9 = 3 i$ , $- 50 = 5 i 2$ . Pull the $- 1$ out as $i$ , then simplify the radical.

Key insight

i^{2} = - 1

is the only definition. Powers cycle every 4 — find the remainder of the exponent mod 4 to read off the answer.

- n = i n

for positive

n

A2.C.2 — Add, subtract, and multiply complex numbers.

Step 01 of 04

A complex number has the form $a + bi$ where $a$ is the real part and $b$ is the imaginary part. Treat $i$ like an algebraic variable — the only special rule is $i^{2} = - 1$ .

Step 02 of 04

Add / subtract. Combine real with real, imaginary with imaginary.

(3 + 5 i) + (2 - 7 i) = 5 - 2 i

(4 + 2 i) - (1 - 3 i) = 3 + 5 i

Step 03 of 04

Multiply. FOIL, then replace $i^{2}$ with $- 1$ .

(2 + 3 i) (4 - i)

= 8 - 2 i + 12 i - 3 i^{2}

= 8 + 10 i - 3 (- 1) = 11 + 10 i

The $i^{2} \to - 1$ step always FLIPS the sign of that piece — easy to forget.

Step 04 of 04

Special product — conjugates. $(a + bi)$ and $(a - bi)$ are complex conjugates. Their product is always REAL:

(a + bi) (a - bi) = a^{2} - (bi)^{2} = a^{2} + b^{2}

Example: $(3 + 4 i) (3 - 4 i) = 9 + 16 = 25$ . This is the trick that powers complex division (next skill).

Key insight

Complex arithmetic = polynomial arithmetic with the rule

i^{2} = - 1

. Add / subtract componentwise. Multiply via FOIL, then replace

i^{2}

. Conjugate pairs multiply to a real number — useful.

A2.C.3 — Divide complex numbers — rationalize the denominator.

Step 01 of 04

You can't have $i$ in a final answer's denominator (it's not "in standard form"). Multiply numerator and denominator by the conjugate of the denominator to clear it.

The conjugate of $a + bi$ is $a - bi$ . Their product is the real number $a^{2} + b^{2}$ .

Step 02 of 04

Worked example. Simplify $\frac{3 + 2 i}{1 - i}$ . Multiply top and bottom by the conjugate $1 + i$ :

\frac{3 + 2 i}{1 - i} \cdot \frac{1 + i}{1 + i} = \frac{( 3 + 2 i ) ( 1 + i )}{( 1 - i ) ( 1 + i )}

Step 03 of 04

Multiply both products.

(3 + 2 i) (1 + i) = 3 + 3 i + 2 i + 2 i^{2} = 3 + 5 i - 2 = 1 + 5 i

(1 - i) (1 + i) = 1 + 1 = 2

So:

\frac{1 + 5 i}{2} = \frac{1}{2} + \frac{5}{2} i

Standard form, no $i$ in the denominator.

Step 04 of 04

Pure imaginary denominator. If the denominator is just $bi$ , multiply by $i / i$ :

\frac{4}{2 i} \cdot \frac{i}{i} = \frac{4 i}{2 i ^{2}} = \frac{4 i}{- 2} = - 2 i

The "conjugate" of $2 i$ is $- 2 i$ (real part is 0), but just multiplying by $i$ works too.

Key insight

Multiply numerator and denominator by the conjugate of the denominator. The denominator becomes a real number (sum of squares); the numerator gets FOILed normally. Result is in standard

a + bi

form.

A2.C.4 — Complex roots of quadratics.

Step 01 of 04

Some quadratics have a NEGATIVE discriminant $(b^{2} - 4 a c < 0)$ — the parabola never crosses the x-axis. These equations don't have real solutions, but they DO have complex solutions that come in conjugate pairs.

The quadratic formula still works. The square root of a negative number just becomes $i$ -scaled.

Step 02 of 04

Worked example. Solve $x^{2} - 4 x + 13 = 0$ .

x = \frac{4 \pm 16 - 52}{2} = \frac{4 \pm - 36}{2}

= \frac{4 \pm 6 i}{2} = 2 \pm 3 i

Two complex solutions: $2 + 3 i$ and $2 - 3 i$ — a conjugate pair. This always happens with real-coefficient quadratics.

Step 03 of 04

Conjugate root theorem. If a polynomial with real coefficients has $a + bi$ as a zero, then $a - bi$ is also a zero. Complex roots ALWAYS come in pairs.

Practical use: if you find one complex root by other means, you get its conjugate for free. Combined, those two roots correspond to a real quadratic factor.

Step 04 of 04

Discriminant decides everything.

$b^{2} - 4 a c > 0$	→two distinct real roots
$b^{2} - 4 a c = 0$	→one real root (double)
$b^{2} - 4 a c < 0$	→two complex conjugate roots

Key insight

When the discriminant is negative, the quadratic formula still works — you just get an

i

in the answer. Real-coefficient polynomials always produce complex roots in CONJUGATE PAIRS, never alone.

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A2.C.1 – A2.C.4 · CCSS N.CN.A.1 / N.CN.A.2 / N.CN.A.3 / N.CN.C.7·Full curriculum →