College Algebra introduced i=\sqrt{-1} so that every polynomial equation could have a solution. There you added, multiplied, and conjugated complex numbers as rules on symbols. This course uses those same rules on nearly every page. But from here on, a complex number is a point in a plane. Each step you take with it is something you can watch happen. Getting these skills back to full speed now clears the way for the pictures in the next section.
Conjugating a number means flipping the sign in front of i and touching nothing else. So 3+4i becomes 3-4i, and the 3 never moves.
Multiplying a number by its own conjugate is the whole trick behind dividing complex numbers. It turns something with a stray i in it into an ordinary real number. It is the same move as multiplying a fraction by \frac{\sqrt2}{\sqrt2} to clear a root from underneath.
Here is the slip people make. Flipping the sign of the whole number gives -z, not \bar z. The little horizontal line drawn over the letter is called an overline, or "bar" for short. Here it is the notation for "the conjugate": same real part, flip the sign on the imaginary part, nothing more. The conjugate only reflects the point across the horizontal axis. It never sends the point to the opposite side of the origin.
Let z=a+bi and w=c+di, with a,b,c,d\in\mathbb R. Addition and subtraction act on each part separately: z\pm w=(a\pm c)+(b\pm d)i. Multiplication works as it does for two binomials, with one extra step: i^2=-1. So zw=(ac-bd)+(ad+bc)i.
Division clears the imaginary part from the denominator. Multiply top and bottom by the denominator's conjugate. For w=c+di\ne 0, the conjugate is \bar w=c-di. The product w\bar w=c^2+d^2 is a nonnegative real number. So \dfrac{z}{w}=\dfrac{z\bar w}{w\bar w} always has a real number underneath.
That same identity gives the modulus its usual formula: |z|=\sqrt{z\bar z}=\sqrt{a^2+b^2}. This is the ordinary distance formula applied to the point (a,b).
Conjugation respects sums and products: \overline{z+w}=\bar z+\bar w and \overline{zw}=\bar z\,\bar w. These two facts do real work once functions of z enter in Chapter 2. There, \overline{f(z)} and f(\bar z) are worth comparing.
One more aside is worth carrying forward. Multiplying by w=c+di can be written as the real 2\times2 matrix \begin{pmatrix}c&-d\\d&c\end{pmatrix} acting on the column (a,b)^T. Matrix multiplication reproduces zw exactly. This is one reason multiplying by a fixed complex number is always a rotation combined with a scaling.
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
Add (5 - 2i) + (-3 + 6i). Enter your answer as a complex number.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Add like with like: real parts 5+(-3)=2, imaginary parts -2i+6i=4i. Result: 2+4i.
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Why
From College Algebra
Multiply (1 + i)(1 - i)(2 + i), entering your answer as a complex number.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Multiply the conjugate pair first: (1+i)(1-i) = 1-i^2 = 1+1 = 2. Then 2(2+i) = 4+2i.
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Why
Multiply the first two factors together first — they're a conjugate pair, so the result is a plain real number.
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
Compute |3 + 4i|, the distance from 3 + 4i to 0 in the complex plane.
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Why
The modulus is \sqrt{a^2 + b^2} = \sqrt{9 + 16} = \sqrt{25} = 5 — Pythagoras on the 3-4-5 triangle. Equivalently, z\bar{z} = (3+4i)(3-4i) = 25, and the modulus is its square root.
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Why
Use the modulus formula, the square root of the sum of the squares of the real and imaginary parts.
Compute 3^2+4^2 before taking the square root.
From College Algebra
Compute |6 - 8i|.
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Why
The modulus is \sqrt{6^2+8^2} = \sqrt{36+64} = \sqrt{100} = 10 — the 6-8-10 Pythagorean triple.
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Why
Use the modulus formula, the square root of the sum of the squares of the real and imaginary parts.
A number and its own conjugate
Drag the point — its conjugate always mirrors it across the real axis, and the two always sum to a purely real number.
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Worked example
All four operations on one pair
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1real parts and imaginary parts combine separately
For z=2+3i and w=1-2i, find z+w, z-w, zw, and z/w. z+w = (2+1)+(3-2)i = 3+i, \qquad z-w = (2-1)+(3+2)i = 1+5i
real parts and imaginary parts combine separately
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2zw = (2+3i)(1-2i) = 2-4i+3i-6i^2 = 2 - i + 6-6i^2 becomes +6 — this is the one step where the algebra stops looking like ordinary binomial multiplication
-6i^2 becomes +6 — this is the one step where the algebra stops looking like ordinary binomial multiplication
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3
Multiply z/w by \bar w/\bar w = (1+2i)/(1+2i) so the denominator becomes real: \frac{z}{w} = \frac{(2+3i)(1+2i)}{(1-2i)(1+2i)} = \frac{2+4i+3i+6i^2}{1+4} = \frac{-4+7i}{5}
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4\boxed{z+w=3+i,\ \ z-w=1+5i,\ \ zw=8-i,\ \ z/w=-\tfrac45+\tfrac75 i}
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Worked example
Conjugate, modulus, and the identity linking them
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1only the sign in front of i flips
For z=-1+2i, find \bar z, |z|, and confirm z\bar z=|z|^2. \bar z = -1-2i
only the sign in front of i flips
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2|z| = \sqrt{(-1)^2+2^2} = \sqrt5
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3z\bar z = (-1+2i)(-1-2i) = 1+2i-2i-4i^2 = 1+4 = 5 = |z|^2
The two computations agree exactly, which is exactly the identity the division rule above leans on. \boxed{\bar z=-1-2i,\quad |z|=\sqrt5}
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Picture three points: the origin, z, and z+w. The length |z+w| is the straight-line distance from the origin to z+w. The length |z| is the distance from the origin to z. And |w| is the distance from z to z+w. That segment is shifted, but it has the same length as the segment from the origin to w.
Those three lengths form a triangle. No side of a triangle is ever longer than the other two sides added together. That is the entire content of the inequality. Nothing about complex numbers changes the ordinary fact you already know about triangles.
The two sides tie in one case: the "triangle" collapses flat. Then z and w point in the exact same direction from the origin, or one of them is 0. The straight path and the detour through z turn out to be secretly the same path.
A second modulus fact this course uses constantly is the triangle inequality. For any z,w\in\mathbb C, |z+w| \le |z|+|w|. Both sides measure a length. The left side, |z+w|, is the distance from the origin to the single point z+w. The right side, |z|+|w|, is the sum of two separate distances, walked one after another. So the claim is about plain distances: going straight from 0 to z+w is never longer than detouring through z first.
Equality holds exactly when z and w lie on a common ray from the origin. That means one of them is 0, or one is a nonnegative real multiple of the other. In that case the detour through z never leaves the straight path.
Apply the same fact to z-w and w in place of z and w. This gives the reverse triangle inequality, \big||z|-|w|\big|\le|z-w|. This course calls on it whenever it needs to control |z| using only a bound on |z-w|.
Theorem
The Triangle Inequality (Complex Numbers)
also known as: the modulus of a sum is at most the sum of the moduli
For complex numbers z,w, |z+w| \le |z| + |w|, with equality if and only if z and w lie on a common ray from the origin — that is, one of z,w is 0, or one is a nonnegative real multiple of the other.
Corollary (reverse triangle inequality). For complex numbers z,w, \big||z|-|w|\big| \le |z-w|.
Main inequality. Write |z+w|^2 using |\zeta|^2 = \zeta\bar\zeta for any complex number \zeta: |z+w|^2 = (z+w)\overline{(z+w)} = (z+w)(\bar z+\bar w) = z\bar z + z\bar w + w\bar z + w\bar w = |z|^2 + \big(z\bar w + \overline{z\bar w}\big) + |w|^2, using w\bar z = \overline{z\bar w}. Since \zeta+\bar\zeta = 2\,\mathrm{Re}(\zeta) for any complex \zeta, the middle term is 2\,\mathrm{Re}(z\bar w), so |z+w|^2 = |z|^2 + 2\,\mathrm{Re}(z\bar w) + |w|^2. For any complex number \zeta, \mathrm{Re}(\zeta) \le |\zeta| (the real part of a number never exceeds its modulus), applied to \zeta = z\bar w: \mathrm{Re}(z\bar w) \le |z\bar w| = |z||\bar w| = |z||w|. Substituting, |z+w|^2 \le |z|^2 + 2|z||w| + |w|^2 = \big(|z|+|w|\big)^2. Both |z+w| and |z|+|w| are nonnegative real numbers, so taking square roots preserves the inequality: |z+w|\le|z|+|w|.
Equality analysis. Equality holds throughout exactly when \mathrm{Re}(z\bar w) = |z||w|, i.e. when \mathrm{Re}(z\bar w) = |z\bar w|. For any complex number \zeta = a+bi, \mathrm{Re}(\zeta) = |\zeta| forces b=0 and a\ge0 — a real part equal to the full modulus can only happen for a nonnegative real number, since otherwise |\zeta| = \sqrt{a^2+b^2} > a = \mathrm{Re}(\zeta) whenever b\neq0 or a<0. So equality holds iff z\bar w is a nonnegative real number. If w=0 this holds trivially (with z\bar w = 0), matching "one of z,w is 0." If w\neq0, z\bar w\ge0 (real) is equivalent to z = \dfrac{z\bar w}{|w|^2}\,w being a nonnegative real multiple of w (multiply both sides by w/|w|^2, using \bar w w = |w|^2) — i.e. z and w point along the same ray from the origin.
Reverse triangle inequality. Apply the main inequality to z-w and w in place of z and w: |z| = |(z-w)+w| \le |z-w|+|w| \implies |z|-|w| \le |z-w|. Symmetrically, applying it to w-z and z: |w| = |(w-z)+z| \le |w-z|+|z| \implies |w|-|z| \le |w-z| = |z-w|, using |w-z|=|{-1}||z-w|=|z-w|. Combining the two, -|z-w| \le |z|-|w| \le |z-w|, which is exactly \big||z|-|w|\big| \le |z-w|.
Worked example
Bounding a sum with the triangle inequality
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1
For |z|=2, find an upper bound on |z^2+3i| without pinning down z exactly. The triangle inequality extends to any sum of two terms: |z^2+3i|\le|z^2|+|3i|
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2
A modulus of a power is the power of the modulus, so |z^2|=|z|^2=4, and |3i|=3: |z^2+3i|\le 4+3=7
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3
That bound holds for every z on the circle |z|=2 at once — the figure shows the specific case z=2, where z^2+3i=4+3i and |4+3i|=5, comfortably under the bound of 7 that holds no matter which point on the circle z actually is. \boxed{|z^2+3i|\le 7 \text{ whenever } |z|=2}
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Where this goes next
The algebra every later proof runs on
Nothing past this section teaches again how to add, multiply, or conjugate a complex number. A proof of the Cauchy-Riemann equations in Chapter 3 leans on this arithmetic. So does a residue computation many chapters from now, and so does a routine limit argument. None of them pauses to reteach it.
A shaky multiplication or a conjugate with the wrong sign here comes back later. It shows up as a wrong answer several chapters on, in a place that looks nothing like this section.
Check your understanding (counts toward your progress)
Why is z\bar z always a real, nonnegative number for z=a+bi?
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Conjugation just flips a sign — it doesn't force positivity on its own. It's the SUM OF SQUARES that guarantees a nonnegative real result here.
z multiplied by itself, z^2, is not generally real. Take (1+i)^2=2i as a counterexample; it's specifically z times its own conjugate that behaves this way.
i^2=-1 is true, but that alone doesn't erase the imaginary part — the product still needs expanding to see the middle terms cancel.
Why
(a+bi)(a-bi)=a^2-(bi)^2=a^2+b^2, a sum of two real squares, which can never be negative.
Multiply (a+bi)(a-bi) out directly and watch the middle terms cancel.
Compute (4-3i)+(2+7i).
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Combining the imaginary parts as -3-7 subtracts instead of adds — (-3)+7=4, not -10.
The real coefficient on the first imaginary term is -3, not 3 — keep its negative sign when combining with 7.
The real parts simply add: 4+2=6. There's no reason to subtract them just because a -3i shows up elsewhere in the problem.
Why
Real parts and imaginary parts combine separately: (4+2)+((-3)+7)i = 6+4i.
Divide (1+3i)/(2-i). Enter your answer as a complex number.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
\dfrac{(1+3i)(2+i)}{(2-i)(2+i)}=\dfrac{2+i+6i+3i^2}{4+1}=\dfrac{-1+7i}{5}.
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Why
Multiply numerator and denominator by the denominator's conjugate, 2+i.
Which expression is equivalent to (a+bi)^2 for real a,b?
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That's |z|^2=z\bar z, a different product. Squaring z itself still keeps an imaginary part, since no conjugate is involved.
Close, but i^2 flips the sign on b^2 — it should be a^2-b^2, not a^2+b^2.
This drops the imaginary part entirely — z^2 keeps one unless b=0.
Why
(a+bi)^2=a^2+2abi+b^2i^2=a^2+2abi-b^2=(a^2-b^2)+2abi.
Expand (a+bi)^2=a^2+2abi+(bi)^2 and simplify (bi)^2.
Practice: Adding, multiplying, and conjugating, one operation at a time (5 problems · doesn’t affect your progress)
Which operation on z=a+bi produces a-bi?
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Negation gives -a-bi — it flips BOTH signs, not just the imaginary one.
Squaring produces (a+bi)^2, a completely different expression with its own real and imaginary parts.
The modulus is a single nonnegative real number, \sqrt{a^2+b^2} — not another complex number.
Why
Conjugation is defined as flipping only the sign in front of i: \overline{a+bi}=a-bi.
Conjugation
Simplify (5+2i)-(3-4i) and express the result as a complex number.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
(5-3)+(2-(-4))i=2+6i.
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Why
Multiply (1+i)(1-i). Enter your answer as a complex number (a real result is fine).
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
(1+i)(1-i)=1-i^2=1+1=2 — a real number, since this is exactly z\bar z=|z|^2.
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Why
This is z\bar z for z=1+i.
Carry out the division (4+2i)/(1+i) and give the quotient as a complex number.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
\dfrac{(4+2i)(1-i)}{(1+i)(1-i)}=\dfrac{4-4i+2i-2i^2}{2}=\dfrac{6-2i}{2}=3-i.
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Why
Multiply top and bottom by the conjugate 1-i.
A student simplifies (3-2i)^2 as 9+4i^2=9-4=5. What's the actual mistake?
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(-2i)^2=-4 was subtracted correctly — the issue isn't that sign, it's a whole missing term.
There's no conjugate involved here to flip a sign on — this is a straight squaring, not a division.
5 is not correct: (3-2i)^2=9-12i+4i^2=9-12i-4=5-12i, and the -12i term never appeared.
Why
(3-2i)^2 expands to 9-2(3)(2i)+(2i)^2=9-12i-4=5-12i. Squaring a binomial needs all three terms, including the cross term.
The middle term, -12i, was dropped entirely