Division & Conjugates
Division & Conjugates
The conjugate of $z = a+bi$ is $\bar{z} = a - bi$.
$$z \cdot \bar{z} = a^2 + b^2 \quad (\text{always a real number})$$
$$|z| = \sqrt{a^2 + b^2}$$
Note: $z \cdot \bar{z} = |z|^2$.
To divide, multiply numerator and denominator by the conjugate of the denominator:
$$\frac{a+bi}{c+di} = \frac{(a+bi)(c-di)}{(c+di)(c-di)} = \frac{(ac+bd)+(bc-ad)i}{c^2+d^2}$$
Find the conjugate and modulus of $z = 3 - 4i$.
$\bar{z} = 3 + 4i$. $|z| = \sqrt{9+16} = 5$.
Compute $\dfrac{2+3i}{1-i}$.
$$\frac{(2+3i)(1+i)}{(1-i)(1+i)} = \frac{2+2i+3i+3i^2}{1+1} = \frac{-1+5i}{2} = -\frac{1}{2} + \frac{5}{2}i$$
Compute $\dfrac{5}{2+i}$.
$$\frac{5(2-i)}{(2+i)(2-i)} = \frac{10-5i}{5} = 2 - i$$
Practice Problems
Show Answer Key
1. $5-2i$
2. $5$
3. $\frac{-i}{1} = -i$
4. $\frac{(4+2i)(1-i)}{2} = \frac{6-2i}{2} = 3-i$
5. $\frac{6(3+2i)}{13} = \frac{18+12i}{13}$
6. $4+25 = 29$
7. $\frac{(1+i)^2}{2} = \frac{2i}{2} = i$
8. $\sqrt{1+1} = \sqrt{2}$
9. $\frac{(3-i)(2-i)}{5} = \frac{5-5i}{5} = 1-i$
10. $4i$
11. $(a+bi)+(a-bi) = 2a = 2\text{Re}(z)$ ✓
12. $\frac{i(2-3i)}{13} = \frac{3+2i}{13}$