The number i and the complex numbers

Every new set of numbers was born from an equation with no solution in the set before it:

• $\displaystyle {x+5}={2}$ has no solution in the natural numbers $\displaystyle \mathbb{N}$; it has one in the integers $\displaystyle \mathbb{Z}$: $\displaystyle {x}={-3}$.
• $\displaystyle {3x}={2}$ has no solution in the integers; it has one in the rationals $\displaystyle \mathbb{Q}$: $\displaystyle {x}={\frac{2}{3}}$.
• $\displaystyle {x^2}={2}$ has no solution in the rationals; it has in the reals $\displaystyle \mathbb{R}$: $\displaystyle {x}={\pm\sqrt{2}}$.
• $\displaystyle {x^2}={-1}$ has no solution in the reals: the square of a real number is never negative.

In the 16th century mathematicians in Italy used roots of negative numbers to solve equations of the third degree, but did not count them as numbers and called them “imaginary”. Only in the 19th century, after Gauss’s work, were they accepted as numbers in every sense.

The number $\displaystyle i$ is a number whose square is $\displaystyle -1$:

A number of the form $\displaystyle {z}={a+bi}$, with $\displaystyle a$ and $\displaystyle b$ real, is called a complex number, and the form $\displaystyle a+bi$ is called its algebraic form.

• $\displaystyle a$ is the real part: $\displaystyle {\operatorname{Re}(z)}={a}$.
• $\displaystyle b$ is the imaginary part: $\displaystyle {\operatorname{Im}(z)}={b}$ – a real number, without the $\displaystyle i$.
• When $\displaystyle {b}={0}$, the number is real: every real number is a complex number too.
• When $\displaystyle {a}={0}$ and $\displaystyle b\neq0$, the number is purely imaginary: $\displaystyle 3i$, $\displaystyle -i$.

For example, in $\displaystyle {z}={4-7i}$: $\displaystyle {\operatorname{Re}(z)}={4}$ and $\displaystyle {\operatorname{Im}(z)}={-7}$.

Both $\displaystyle {(3i)^2}={9i^2}={-9}$ and $\displaystyle {(-3i)^2}={-9}$. The root $\displaystyle \sqrt{-9}$ is one of them, $\displaystyle 3i$:

• $\displaystyle {\sqrt{-16}}={4i}$, $\displaystyle {\sqrt{-5}}={i\sqrt{5}}$.
• Careful: $\displaystyle {\sqrt{a}\cdot\sqrt{b}}={\sqrt{ab}}$ only when $\displaystyle a,b\geq0$. $\displaystyle \sqrt{-4}\cdot\sqrt{-9}$ is not $\displaystyle {\sqrt{36}}={6}$ but $\displaystyle {2i\cdot3i}={6i^2}={-6}$.

The powers of $\displaystyle i$ repeat every four:

• $\displaystyle {i^1}={i}$, $\displaystyle {i^2}={-1}$, $\displaystyle {i^3}={i^2\cdot i}={-i}$, $\displaystyle {i^4}={(i^2)^2}={1}$.
• $\displaystyle {i^5}={i^4\cdot i}={i}$, and from there it all repeats.

So $\displaystyle i^n$ depends only on the remainder of $\displaystyle n$ divided by 4. The example below works out $\displaystyle i^{2026}$:

• $\displaystyle {2026}={4\cdot506+2}$.
• $\displaystyle {i^{2026}}={(i^4)^{506}\cdot i^2}={1\cdot(-1)}={-1}$.

Two complex numbers are equal when their real parts are equal and their imaginary parts are equal:

One equation in complex numbers is therefore two equations in real numbers. The example below finds real $\displaystyle x$ and $\displaystyle y$ with $\displaystyle {(x+2y)+(x-y)i}={5+2i}$:

• the real parts: $\displaystyle {x+2y}={5}$;
• the imaginary parts: $\displaystyle {x-y}={2}$;
• subtracting the equations: $\displaystyle {3y}={3}$, so $\displaystyle {y}={1}$ and $\displaystyle {x}={3}$.

Calculate, and write the result in algebraic form.

Find real $\displaystyle x$ and $\displaystyle y$ that satisfy the equality.
Write the two equations – of the real parts and of the imaginary parts – in the first line, and solve.

$\displaystyle {(2x-y)+(x+y)i}={4+5i}$

$\displaystyle {(x+y)+(3x-y)i}={2-6i}$

Given the number $\displaystyle {z}={(x-3)+(x^2-4)i}$, with $\displaystyle x$ real.

Find the values of $\displaystyle x$ for which $\displaystyle z$ is real: solve the equation $\displaystyle {\operatorname{Im}(z)}={0}$.

Find the value of $\displaystyle x$ for which $\displaystyle z$ is purely imaginary. Explain why you must check the imaginary part too.