# ARITHMETIC WITH COMPLEX NUMBERS

• PRACTICE (online exercises and printable worksheets)

Whenever you get a new mathematical object (like complex numbers), you need to develop tools to work with the new object.

Arithmetic with complex numbers was introduced in a prior lesson:   Arithmetic with Complex Numbers in the Algebra II materials. Quickly read through this earlier section and do plenty of the online exercises to make sure that you understand all the concepts there.

For your convenience, this current lesson gives a quick summary of important concepts from the earlier lesson, and also discusses two new topics: the complex conjugate, and division of complex numbers.

 ARITHMETIC WITH COMPLEX NUMBERS adding, subtracting, multiplying, dividing, complex conjugate, more Let $\,a\,$, $\,b\,$, $\,c\,$, and $\ d\$ be real numbers, and let $\,i:=\sqrt{-1}\,$. addition and subtraction: $$\begin{gather} (a+bi)\, + \, (c+di)\ =\ (a+c) + (b+d)i\cr (a+bi)\, - \, (c+di)\ =\ (a-c) + (b-d)i \end{gather}$$ To add complex numbers: add the real parts; this becomes the real part of the sum add the imaginary parts; this becomes the imaginary part of the sum The same idea holds for differences. You can also just think of combining like terms—the real number terms and the $\,i\,$ terms. multiplication: $$(a+bi)(c+di) \ =\ (ac-bd) + (ad+bc)i$$ You certainly don't want to memorize this formula. Just use FOIL to multiply things out! \begin{alignat}{2} (a+bi)(c+di) &= \overbrace{\ ac\ \strut}^{\text{F}} + \overbrace{\ adi\ \strut}^{\text{O}} + \overbrace{\ bci\ \strut}^{\text{I}} + \overbrace{\ bdi^2\ \strut}^{\text{L}} && \qquad\text{using FOIL}\cr &= (ac + bdi^2) + (adi + bci)&& \qquad\text{re-group}\cr &= (ac - bd) + (ad + bc)i && \qquad i^2 = -1 \text{ and factoring} \end{alignat} multiplication by a real number A special case of multiplication (above) is multiplication by a real number. For all real numbers $\,k\$: $$k(a + bi) = ka + kbi$$ Notice that both the real and imaginary parts get multiplied by $\,k\,$. the complex conjugate: By definition, the complex conjugate of $\,a$ $+\,bi\,$ is $\,a$ $-\,bi\,$. (Hover over the yellow with your cursor.) Note that the real part stays the same; you take the opposite of the imaginary part. The complex conjugate of $\,a + bi\,$ is notated by $\overline{a + bi}\,$, so: $$\overline{a + bi} = a - bi$$ When a complex number is multiplied by its conjugate, a nonnegative real number results. Letting $\,z = a + bi\,$: $$z\overline{z} = (a+bi)(a-bi) = a^2 - abi + abi - b^2i^2 = a^2 + b^2$$ division: For division, we need the additional requirement that $\,c\,$ and $\,d\,$ are not both zero. $$\frac{a+bi}{c+di} \ =\ \left(\frac{ac + bd}{c^2 + d^2}\right) + \left(\frac{bc - ad}{c^2 + d^2}\right)i$$ Again, you certainly don't want to memorize this formula. Just multiply by $\,1\,$ in the form of the complex conjugate of the denominator over itself, and then simplify: \begin{alignat}{2} \frac{a+bi}{c+di} &= \frac{a+bi}{c+di}\cdot\overbrace{\frac{c-di}{c-di}}^{= 1} &&\qquad \text{multiply by \,1\,}\cr\cr &= \frac{(a+bi)(c-di)}{(c+di)(c-di)} &&\qquad \text{multiply across}\cr\cr &= \frac{(ac+ bd) + (bc-ad)i}{c^2 + d^2}&&\qquad \text{multiply out numerator and denominator}\cr\cr &= \left(\frac{ac + bd}{c^2 + d^2}\right) + \left(\frac{bc - ad}{c^2 + d^2}\right)i \end{alignat} naming conventions for complex numbers: Two common variable names for complex numbers are $\,z\,$ and $\,w\,$.

## EXAMPLES

Let $z = 2 - 3i\,$ and $\,w = -5 + 7i\,$.
Then: \begin{alignat}{2} z + w \ &= (2-3i) + (-5 + 7i) \ =\ (2-5) + (-3 + 7)i \ =\ -3 + 4i\cr z - w \ &= (2-3i) - (-5 + 7i) \ =\ (2+5) + (-3 - 7)i \ =\ 7 - 10i\cr\cr zw\ &= (2 - 3i)(-5 + 7i) \ =\ -10 + 14i + 15i -21i^2 \ =\ (-10 + 21) + 29i \ =\ 11 + 29i\cr\cr \frac{z}{w}\ &= \frac{2-3i}{-5 + 7i}\cdot\frac{-5-7i}{-5-7i}\ =\ \frac{(2-3i)(-5-7i)}{25 + 49}\ = \frac{(-10-21) + (-15-14)i}{74}\ =\ \frac{-31 - 29i}{74}\ =\ -\frac{31}{74} - \frac{29}{74}i\cr\cr\cr \overline{z}\ &= \overline{2-3i} \ =\ 2 + 31\cr\cr \overline{w}\ &= \overline{-5 + 7i}\ =\ -5 - 7i\cr\cr z\overline{z}\ &= (2-3i)(2+3i) \ =\ 4 + 6i - 6i - 9i^2\ =\ 4 + 9 \ =\ 13\cr\cr -4z + \overline{2w}\ &= -4(2 - 3i) + \overline{2(-5 + 7i)}\ =\ -8 + 12i + (\overline{-10 + 14i})\ =\ -8 + 12i + (- 10 - 14i)\ =\ -18 - 2i \end{alignat}

Master the ideas from this section