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The Conjugate

Part of the Quadratics & Complex Numbers section of Coddy's Algebra 2 journey. Lesson 21 of 48.

The conjugate of a+bia+bi is a−bia-bi: the same real part, with the sign of the imaginary part flipped. The conjugate of 2+7i2+7i is 2−7i2-7i, and the conjugate of 6−i6-i is 6+i6+i.

Multiply a number by its conjugate: (2+7i)(2−7i)=4−14i+14i−49i2\left(2+7i\right)\left(2-7i\right) = 4-14i+14i-49{i}^{2}. The two i terms are opposites and cancel, and −49i2-49{i}^{2} is +49. The result is 4 + 49, which is 53, a real number.

For real a and b that always happens: (a+bi)(a−bi)=a2+b2\left(a+bi\right)\left(a-bi\right) = {a}^{2}+{b}^{2}. Compare (a+b)(a−b)=a2−b2\left(a+b\right)\left(a-b\right) = {a}^{2}-{b}^{2} from Algebra: the minus turns into a plus because i2=−1{i}^{2} = -1. This is the tool for the next lesson, since multiplying by the conjugate turns a complex number into a real one.

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Challenge

Multiply (4+3i)(4−3i)\left(4+3i\right)\left(4-3i\right). The answer is a real number.

Try it yourself

(4+3i)(4−3i)\left(4+3i\right)\left(4-3i\right)

Expand: multiply everything out

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