Math 2–3 Honors
Math 2–3 Honors · Lesson Review · October 20, 2026

Complex Numbers and Arithmetic

Meet the imaginary unit, plot complex numbers, add and subtract like terms, multiply using i²=−1, divide with conjugates, and find and check complex quadratic zeros.

31-minute edited lesson13 chaptersCaptions + transcript

01 · Watch

Lesson video

VTT captionsSRT captions
0:0001Why introduce i?2:2402A plane for complex numbers7:0203How the number families fit together10:5204Add and subtract like terms13:0105Multiply and simplify i squared17:2706Conjugate products19:2907The conjugate identity20:4408Divide using a conjugate22:5709A second quotient25:1510Find complex quadratic zeros26:4811Check the complete root28:0512A glimpse of complex functions29:4813Why complex zeros matter

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02 · Understand

Learning targets

03 · Connect

Key ideas

The horizontal coordinate is the real coefficient a; the vertical coordinate is the imaginary coefficient b. The term bi is the imaginary term. Real numbers are included in the complex numbers by taking b=0.

01The imaginary unit
i2=−1,x2=−1 ⟹ x=±ii^2=-1,\qquad x^2=-1\ \Longrightarrow\ x=\pm iNo real number squares to −1. The equation has two complex solutions, i and −i. The symbol i denotes the chosen imaginary unit, not both solutions at once.
02Read a point in the complex plane
a+bi⟷(a,b),2+3i⟷(2,3)a+bi\longleftrightarrow(a,b),\qquad 2+3i\longleftrightarrow(2,3)Move a units horizontally and b units vertically. A complex number is real exactly when b=0; it is nonzero pure imaginary when a=0 and b≠0.
03Combine matching terms
(a+bi)+(c+di)=(a+c)+(b+d)i(a+bi)+(c+di)=(a+c)+(b+d)iCombine real terms with real terms and imaginary terms with imaginary terms. For subtraction, first distribute the negative sign across the entire subtracted expression.
04Multiply, then simplify i²
(a+bi)(c+di)=(ac−bd)+(ad+bc)i(a+bi)(c+di)=(ac-bd)+(ad+bc)iThe usual distributive property still works. The final product bd i² contributes −bd to the real part.
05Conjugates produce a real product
a+bi‾=a−bi,(a+bi)(a−bi)=a2+b2\overline{a+bi}=a-bi,\qquad(a+bi)(a-bi)=a^2+b^2Only the sign of the imaginary term changes. The two middle terms cancel. The product is positive for every nonzero a+bi.
06Divide using the denominator’s conjugate
a+bic+di=(a+bi)(c−di)c2+d2,c+di≠0\frac{a+bi}{c+di}=\frac{(a+bi)(c-di)}{c^2+d^2},\qquad c+di\ne0Multiply both numerator and denominator by the same conjugate. Expand the numerator, simplify i², and write the result in a+bi form.
07Negative discriminants give complex zeros
x=−b±b2−4ac2a,−u=iu  (u≥0)x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},\qquad \sqrt{-u}=i\sqrt{u}\ \ (u\ge0)For a quadratic with real coefficients, nonreal zeros come in conjugate pairs. They do not create real x-intercepts. The product rule for square roots cannot be used indiscriminately on negative real radicands. A zero is an input r with f(r)=0. Every quadratic has two complex zeros when repeated zeros are counted with multiplicity.

04 · Apply

Worked examples

These examples follow the methods developed in the recorded lesson.

Classroom example

Add in standard complex form

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

Group the real coefficients and the imaginary coefficients.

=−2+8i=-2+8i

The result is one complex number, written in a+bi form.

Classroom example

Distribute a subtraction sign

4i−(7+3i)+(4−6i)=4i−7−3i+4−6i4i-(7+3i)+(4-6i)=4i-7-3i+4-6i

The minus sign changes both terms inside the first parentheses.

=−3−5i=-3-5i

The real terms sum to −3 and the imaginary coefficients sum to −5.

Classroom example

Multiply using i²=−1

(2i)(5i)=10i2=−10(2i)(5i)=10i^2=-10

Two imaginary factors can have a real product.

(2+3i)(4−5i)=8−10i+12i−15i2(2+3i)(4-5i)=8-10i+12i-15i^2

Distribute all four products.

=8+2i+15=23+2i=8+2i+15=23+2i

Replace i² by −1 before combining real terms.

Classroom example

Why conjugates help

(2+3i)(2−3i)=4−6i+6i−9i2=13(2+3i)(2-3i)=4-6i+6i-9i^2=13

The imaginary middle terms cancel.

(a+bi)(a−bi)=a2−abi+abi−b2i2=a2+b2(a+bi)(a-bi)=a^2-abi+abi-b^2i^2=a^2+b^2

The same cancellation works for any real a and b.

Classroom example

Divide by 2−5i

3i2−5i⋅2+5i2+5i=6i+15i24+25\frac{3i}{2-5i}\cdot\frac{2+5i}{2+5i}=\frac{6i+15i^2}{4+25}

Use the conjugate of the denominator.

=−15+6i29=−1529+629i=\frac{-15+6i}{29}=-\frac{15}{29}+\frac6{29}i

The real denominator is 2²+5²=29.

Classroom example

A quotient that simplifies to halves

1+2i3+i⋅3−i3−i=3−i+6i−2i29−i2\frac{1+2i}{3+i}\cdot\frac{3-i}{3-i}=\frac{3-i+6i-2i^2}{9-i^2}

Multiply both numerator and denominator by 3−i.

=5+5i10=12+12i=\frac{5+5i}{10}=\frac12+\frac12i

Divide both terms by the real denominator.

Classroom example

Complex zeros and complex factors

f(x)=x2+2x+2,Δ=22−4(1)(2)=−4f(x)=x^2+2x+2,\qquad \Delta=2^2-4(1)(2)=-4

This version of the classroom quadratic has no real zeros.

x=−2±−42=−2±2i2=−1±ix=\frac{-2\pm\sqrt{-4}}2=\frac{-2\pm2i}2=-1\pm i

Keep both signs and simplify the entire numerator.

f(x)=(x+1−i)(x+1+i)=(x+1)2+1f(x)=(x+1-i)(x+1+i)=(x+1)^2+1

Use x minus each zero; expanding checks the factorization.

(−1−i)2=1+2i+i2=2i,f(−1−i)=2i−2−2i+2=0(-1-i)^2=1+2i+i^2=2i,\qquad f(-1-i)=2i-2-2i+2=0

Check the entire root −1−i. The real part is negative one; a temporary sign on the board is corrected during the discussion.

05 · Avoid

Common mistakes

Replacing i² with +1.

The defining identity is i²=−1. For example, −15i² becomes +15.

Forgetting the second subtraction sign.

−(7+3i)=−7−3i. Distribute the minus sign to every term.

Treating every complex number as nonreal.

Every real number is also complex: a=a+0i. A nonreal complex number has a nonzero imaginary coefficient.

Using the numerator’s conjugate.

For division, use the denominator’s conjugate and multiply both top and bottom by it.

Calling a complex zero an x-intercept.

The real coordinate graph only shows real zeros as x-intercepts. Nonreal zeros are values in the complex number system.

Changing the quadratic’s constant unnoticed.

x²+2x−1 has real zeros −1±√2. Changing the constant to +2 gives x²+2x+2 with complex zeros −1±i. Check the exact equation before applying the formula.

06 · Check yourself

Try it first

Use these five fresh problems to check your understanding. Open a solution after you try it.

1Plot z, identify its real and imaginary coefficients and conjugate, then add w. z=−3+2i,w=5−7iz=-3+2i,\qquad w=5-7i

z⟷(−3,2),Re⁡(z)=−3,Im⁡(z)=2,zˉ=−3−2iz\longleftrightarrow(-3,2),\quad\operatorname{Re}(z)=-3,\quad\operatorname{Im}(z)=2,\quad\bar z=-3-2i

Move left 3 and up 2. The imaginary coefficient is 2; the imaginary term is 2i.

z+w=(−3+5)+(2−7)i=2−5iz+w=(-3+5)+(2-7)i=2-5i

Combine like terms.

2Subtract and write the result in standard complex form. (6−i)−(2+4i)(6-i)-(2+4i)

6−i−2−4i6-i-2-4i

Distribute the outside negative to both terms.

=4−5i=4-5i

Combine the real and imaginary terms separately.

3Multiply and simplify. (3+2i)(1−4i)(3+2i)(1-4i)

3−12i+2i−8i23-12i+2i-8i^2

Distribute all four products.

=3−10i+8=11−10i=3-10i+8=11-10i

Use i²=−1.

4Divide and write the result in a+bi form. 2+3i1−2i\frac{2+3i}{1-2i}

(2+3i)(1+2i)(1−2i)(1+2i)=2+4i+3i+6i21+4\frac{(2+3i)(1+2i)}{(1-2i)(1+2i)}=\frac{2+4i+3i+6i^2}{1+4}

Multiply by the denominator’s conjugate.

=−4+7i5=−45+75i=\frac{-4+7i}{5}=-\frac45+\frac75i

The denominator is 5, so division is valid.

5Find both complex zeros and give a factorization over the complex numbers. p(x)=x2−6x+13p(x)=x^2-6x+13

Δ=(−6)2−4(1)(13)=−16\Delta=(-6)^2-4(1)(13)=-16

The negative discriminant rules out real zeros.

x=6±−162=6±4i2=3±2ix=\frac{6\pm\sqrt{-16}}2=\frac{6\pm4i}2=3\pm2i

The zeros are a conjugate pair.

p(x)=(x−3−2i)(x−3+2i)=(x−3)2+4p(x)=(x-3-2i)(x-3+2i)=(x-3)^2+4

Expanding gives x²−6x+13 again; the real graph stays above the x-axis.

07 · Revisit

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