Complex Numbers: Key Concepts
- The complex number refers to an entity of the form $a+bi$, where $a$ and $b$ are ordinary real numbers, and $i$ has the property that $i^2=-1$.
- The solution to the equation $x^2 = -4$ is $x = 2i$ and $x=-2i$.
- If $z = a + bi$, where $a$ and $b$ are real numbers, then $a$ is called the
real part of $z$ and $b$ is called the imaginary part of $z$.
- Addition and subtraction of complex numbers: Suppose $a, b, c,$ and $d$ are real numbers. Then
$$ (a + bi) + (c + di) = (a + c) + (b + d)i $$
$$ (a + bi) – (c + di) = (a – c) + (b – d)i $$
- Multiplication of complex numbers:
Suppose $a, b, c$, and $$$ are real numbers. Then
$$ (a + bi)(c + di) = (ac – bd) + (ad + bc)i $$
- Complex conjugate: Suppose $a$ and $b$ are real numbers. The complex conjugate of the complex number $a+bi$ is simply $a – bi$, where the sign between the two terms has been reversed. For example, the complex conjugate of $-2+3i$ is $-2-3i$.
The product of a complex number and its complex conjugate is always a real number. For example,
$$ (2 + i)(2 – i) = (2)^2 – (i)^2 = 4 – (-1) = 5 $$
This is because of the difference of squares identity, which tells us that
$$(x+y)(x-y) = x^2 – y^2$$
If we multiply a complex number $a+bi$ with its complex conjugate $a-bi$, we obtain
$$(a+bi)(a-bi) = a^2 – (bi)^2 = a^2 – b^2i^2 = a^2 + b^2 \quad \textrm{Note: } i^2=-1$$
The expression $a^2+b^2$ is always a real number because both $a$ and $b$ are real.
- Dividing Complex Numbers: To divide complex numbers, we must first eliminate all imaginary numbers from the denominator. This can be done by multiplying the complex number in the denominator by its conjugate. The example below illustrates this procedure:
$$ \dfrac{3+i}{1+3i} = \dfrac{3+i}{1+3i} \times \dfrac{1-3i}{1-3i} = \dfrac{3 – 9i + i – 3i^2}{1 – 9i^2} = \dfrac{6-8i}{10} =
\dfrac{3}{5} – \dfrac{4i}{5} $$
real part of $z$ and $b$ is called the imaginary part of $z$.
$$ (a + bi) + (c + di) = (a + c) + (b + d)i $$
$$ (a + bi) – (c + di) = (a – c) + (b – d)i $$
Suppose $a, b, c$, and $$$ are real numbers. Then
$$ (a + bi)(c + di) = (ac – bd) + (ad + bc)i $$
The product of a complex number and its complex conjugate is always a real number. For example,
$$ (2 + i)(2 – i) = (2)^2 – (i)^2 = 4 – (-1) = 5 $$
This is because of the difference of squares identity, which tells us that
$$(x+y)(x-y) = x^2 – y^2$$
If we multiply a complex number $a+bi$ with its complex conjugate $a-bi$, we obtain
$$(a+bi)(a-bi) = a^2 – (bi)^2 = a^2 – b^2i^2 = a^2 + b^2 \quad \textrm{Note: } i^2=-1$$
The expression $a^2+b^2$ is always a real number because both $a$ and $b$ are real.
$$ \dfrac{3+i}{1+3i} = \dfrac{3+i}{1+3i} \times \dfrac{1-3i}{1-3i} = \dfrac{3 – 9i + i – 3i^2}{1 – 9i^2} = \dfrac{6-8i}{10} =
\dfrac{3}{5} – \dfrac{4i}{5} $$
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