Complex Conjugates

Flip the sign of the imaginary part

The conjugate of 3 + 2i is 3 − 2i. The real part stays as it is, and the imaginary part changes sign.

So the conjugate of 4 − 5i is 4 + 5i, and the conjugate of −1 + i is −1 − i. The real part keeps its sign, whatever it is. A real number such as 7 is its own conjugate, since its imaginary part is 0, and the conjugate of 6i is −6i. Taking the conjugate twice gives back the number you started with.

A number and its conjugate have the same real part and opposite imaginary parts. Added, the imaginary parts cancel and the sum is the real number 6. Subtracted, the real parts cancel and the difference is 4i.

The sum is real

Adding a number to its conjugate cancels the imaginary parts: (3 + 2i) + (3 − 2i) = 6. In general (a + bi) + (a − bi) = 2a, twice the real part.

Subtracting cancels the real parts instead: (3 + 2i) − (3 − 2i) = 4i, and in general the difference is 2bi.

The product is real

Multiply the pair out: $\left(3+2i\right)\left(3-2i\right)=9-6i+6i-4i^2$(3 + 2i)(3 − 2i) = 9 − 6i + 6i − 4i².

The two middle terms, −6i and +6i, cancel. The last term is $-4i^2=-4\times \left(-1\right)=+4$−4i² = −4 × (−1) = +4. So the product is 9 + 4 = 13, a real number with no imaginary part left at all.

The four products of (3 + 2i)(3 − 2i). The two gold pieces, 6i and −6i, cancel. The last piece is $-4i^2=+4$−4i² = +4, so the product is 9 + 4 = 13.

Always $a^2+b^2$a² + b²

For any a + bi: $\left(a+\text{bi}\right)\left(a-\text{bi}\right)=a^2-$(a + bi)(a − bi) = a² − abi + abi $-b^2{i}^2=a^2+b^2$− b²i² = a² + b².

It looks like a difference of two squares, $\left(x+y\right)\left(x-y\right)=x^2-y^2$(x + y)(x − y) = x² − y², and it is one, with y = bi. But $y^2={\left(\text{bi}\right)}^2=b^2{i}^2=-b^2$y² = (bi)² = b²i² = −b², so $x^2-y^2=a^2-\left(-b^2\right)=a^2+b^2$x² − y² = a² − (−b²) = a² + b². The minus sign turns into a plus.

So (1 + 2i)(1 − 2i) = 1 + 4 = 5, (5 − 3i)(5 + 3i) = 25 + 9 = 34 and (4 + 4i)(4 − 4i) = 16 + 16 = 32. The product of a number and its conjugate is never negative, and it is 0 only when a and b are both 0.

A sum of squares factors

Read backward, the same product factors a sum of two squares, which has no real factors: $x^2+9=\left(x+3i\right)\left(x-3i\right)$x² + 9 = (x + 3i)(x − 3i). Check by expanding: the cross terms cancel, and $x^2-{\left(3i\right)}^2=x^2-9i^2=x^2+9$x² − (3i)² = x² − 9i² = x² + 9.

So $x^2+9=0$x² + 9 = 0 when x = 3i or x = −3i, a number and its conjugate. Complex Roots of a Quadratic finds pairs like this from the quadratic formula.

Conjugates go through sums and products

The conjugate of a product is the product of the conjugates. (2 + 3i)(1 + 4i) = −10 + 11i, and $\left(2-3i\right)\left(1-4i\right)=2-8i-3i+12i^2=-10-11i$(2 − 3i)(1 − 4i) = 2 − 8i − 3i + 12i² = −10 − 11i, the conjugate of −10 + 11i.

The reason: taking the conjugate puts −i wherever i was, and ${\left(-i\right)}^2=-1$(−i)² = −1 as well, so every step of the working is the same with every i-term negated. Sums behave the same way. This is why the non-real roots of an equation with real coefficients come in conjugate pairs.

The usual mistakes

Using $a^2-b^2$a² − b². (3 + 2i)(3 − 2i) is not 9 − 4 = 5: the term $-4i^2$−4i² is +4, so the product is 13.

Keeping an i in the answer. (3 + 2i)(3 − 2i) is not 9 + 4i: the cross terms −6i and +6i cancel, and nothing imaginary is left.

Changing the wrong sign. The conjugate of 3 + 2i is not −3 + 2i, and not −3 − 2i, which is −(3 + 2i). Only the sign of the imaginary part changes.