1. $i = \sqrt { - 1}$, ${i^2} = - 1$
2. A complex number is any number of the form a + bi where a and b are real numbers.
To add or subtract two complex numbers, you add or subtract the real parts and the imaginary parts.
(a + bi) + (c + id) = (a + c) + (b + d)i.
(a + bi) - (c + id) = (a - c) + (b - d)i.
Example 1:
(3 - 5i) + (6 + 7i) = (3 + 6) + (-5 + 7)i = 9 + 2i.
(3 - 5i) - (6 + 7i) = (3 - 6) + (-5 - 7)i = -3 - 12i.
Level 1
Level 2
Example 2:
Let's take specific complex numbers to multiply, say 2 + 3i and 2 - 5i.
(2 + 3i)(2 - 5i) = 4 - 10i + 6i - 15i2 = 4 - 4i - 15i2
The definition of i tells us that i2 = -1 . Therefore,
(2 + 3i)(2 - 5i) = 4 - 4i -15(-1) = 19 - 4i.
If you generalize this example, you'll get the general rule for multiplication
(x + yi)(u + vi) = (xu - yv) + (xv + yu)i
Level 1
Level 2
We define the conjugate of a + bi as $\overline {a + bi} = a - bi$
Example 3: $\overline {2 + 3i} = 2 - 3i$
Conjugates are important because a complex number times its conjugate is a real number.
Example 4: $(3 + 4i) \cdot (3 - 4i) = 9 - 12i + 12i - 16{i^2} = 9 - 16 \cdot ( - 1) = 25$
We define modulus or absolute value of complex number a + bi as $\sqrt {{a^2} + {b^2}}$. We write modulus of a + bi as |a + bi|.
Example 4:
${\rm{|3 + 4i|}} = \sqrt {{3^2} + {4^2}} = 5$
Level 1
Level 2
The process of division of complex numbers:
step 1: Find the conjugate of a denominator.
step 2: Multiply the complex fraction, both top and bottom complex number.
Here is the complete division problem:
$$ \begin{align} \frac{{2 + 3i}}{{4 - 5i}} &= \frac{{2 + 3i}}{{4 - 5i}} \cdot \frac{{4 + 5i}}{{4 + 5i}} = \\ &= \frac{{8 + 10i + 12i + 15{i^2}}}{{16 + 20i - 20i - 25{i^2}}} = \\ &= \frac{{8 + 22i + 15 \cdot ( - 1)}}{{16 - 25 \cdot ( - 1)}} = \\ &= \frac{{ - 7 + 22i}}{{41}} = \\ &= - \frac{7}{{41}} + \frac{{22}}{{41}}i \end{align} $$Now, we can write down a general formula for division of complex numbers
$$\frac{{a + bi}}{{c - di}} = \frac{{ac + bd}}{{{c^2} + {d^2}}} + \frac{{bc - ad}}{{{c^2} + {d^2}}}$$Level 1
Level 2
Every nth - order polynomial possess exactly n complex roots.