Q.If $z_1=3,\ z_2=-7i$, and $z_3=5+4i$, show that

Complex Number Arithmetic: A First Look

Imagine you're trying to solve $x^2 + 1 = 0$. You know that no real number squared gives $-1$. The square of any real number is either zero or positive. So this equation has no real solution. But what if we invent a number whose square is $-1$? That's exactly what mathematicians did — and that invention is the imaginary unit $i$, defined by:

A complex number is any number of the form $a + bi$, where $a$ and $b$ are real numbers. Here $a$ is called the real part, and $b$ is called the imaginary part. For example, $3 + 4i$ has real part $3$ and imaginary part $4$.

Why Bother?

Complex numbers let you solve equations that real numbers can't. Every polynomial equation — no matter how complicated — has a solution in the complex numbers. This is the Fundamental Theorem of Algebra, and it's one of the most important results in mathematics.

Arithmetic Operations

The rules are straightforward: treat $i$ like a variable, but remember that $i^2 = -1$.

Addition and Subtraction

Add (or subtract) real parts with real parts, imaginary parts with imaginary parts.

Example: $(2 + 3i) + (4 - 5i) = (2 + 4) + (3 - 5)i = 6 - 2i$

Multiplication

Multiply like binomials, then replace $i^2$ with $-1$.

Example: $(2 + 3i)(4 - 5i) = 2(4) + 2(-5i) + 3i(4) + 3i(-5i)$

$= 8 - 10i + 12i - 15i^2$

$= 8 + 2i - 15(-1)$

$= 8 + 2i + 15 = 23 + 2i$

Division

Division is trickier. The key idea: multiply numerator and denominator by the complex conjugate of the denominator.

The complex conjugate of $a + bi$ is $a - bi$. When you multiply a complex number by its conjugate, you get a real number:

So to divide:

Example: $\frac{2 + 3i}{4 - 5i} = \frac{(2 + 3i)(4 + 5i)}{(4 - 5i)(4 + 5i)} = \frac{8 + 10i + 12i + 15i^2}{16 + 25} = \frac{8 + 22i - 15}{41} = \frac{-7 + 22i}{41} = -\frac{7}{41} + \frac{22}{41}i$

The Big Picture …