Complex numbers

Why a new number is needed. Squaring any real number — positive, negative or zero — always gives a result that is zero or positive. So $x^2=-1$ has no real solution: it sits entirely outside what the real line can offer. Instead of stopping there, a new quantity $i$ (the imaginary unit) is introduced by definition:

where: $i$ is a symbol, not a variable that can take different values — it is fixed once and for all so that squaring it gives $-1$.

The standard form. A complex number is any expression

where: $a=\operatorname{Re}(z)$ is the real part and $b=\operatorname{Im}(z)$ is the imaginary part; the set of all such $z$ is written $\mathbb{C}$. If $b=0$, $z=a$ is an ordinary real number, so $\mathbb{R}\subset\mathbb{C}$ — every real number is complex, but not every complex number is real. If $a=0$ and $b\neq0$, $z=ib$ is called purely imaginary.

Equality of complex numbers. Two complex numbers $a+ib$ and $c+id$ are equal if and only if $a=c$ and $b=d$ — real parts must match real parts, and imaginary parts must match imaginary parts, independently. This single rule is the workhorse used to solve for unknowns tucked inside a complex equation, as the worked example below shows.

Why enlarging the number system is a familiar move. This is not the first time mathematics has hit a wall and grown past it. Natural numbers could not solve $x+5=2$, so negative integers were introduced; integers could not solve $2x=3$, so rational numbers were introduced; rationals could not give an exact value for $\sqrt2$, so irrational and hence real numbers were introduced. Complex numbers are the next, and — for solving polynomial equations — the final step in this chain: once $i$ is admitted, no further enlargement is ever needed to solve a polynomial equation with real or complex coefficients, a deeper fact called the Fundamental Theorem of Algebra, stated here without proof. This single definition of $i$ is therefore the opening move of the whole chapter: everything that follows — the algebra of complex numbers, their modulus and conjugate, the Argand plane, and solving quadratics with a negative discriminant — builds directly on the standard form $a+ib$ introduced here.

Powers of $i$ cycle with period 4. Starting from the definition,

and then the block repeats forever: $i^5=i$, $i^6=-1$, $i^7=-i$, $i^8=1$, and so on. Only the remainder when the exponent is divided by 4 matters, which is exactly what the derivation below makes precise. The figure below places a purely imaginary number and a general complex number on the plane that complex numbers naturally live in — the real part along a horizontal axis, the imaginary part along a vertical one, the same picture the chapter later formalises as the Argand plane.

Problem. Simplify $i^{37}$ and write $3+i^{37}$ in standard form $a+ib$. Then find real numbers $x,y$ such that $(x+2)+i(y-3)=5+7i$.

Step 1. Reduce the exponent using the remainder rule just derived: $37=4\times9+1$, so the remainder is $r=1$, and therefore $i^{37}=i^1=i$.

Step 2. Substitute this back: $3+i^{37}=3+i$, which is already in standard form $a+ib$ with $a=3$ and $b=1$.

Step 3. For the second part, use equality of complex numbers: the real parts must match, so $x+2=5$, giving $x=3$; the imaginary parts must match, so $y-3=7$, giving $y=10$.

Step 4. Check both results together. $3+i^{37}=3+i$, and $(3+2)+i(10-3)=5+7i$, which matches the right-hand side exactly. Final answers: $3+i^{37}=3+i$, and $x=3,\ y=10$.

Competitive stretch

Find $x,y$ if $2x+i(3y-1)=8+11i$. Comparing real parts, $2x=8$ so $x=4$; comparing imaginary parts, $3y-1=11$ so $3y=12$ and $y=4$. The same two-step habit — separate the equation into a real-part equation and an imaginary-part equation, then solve each like an ordinary linear equation — handles every problem of this type, regardless of how the unknowns are arranged inside $z$.

• Writing $\sqrt{-4}=-2i$ or dropping the $i$ entirely — always factor as $\sqrt{-4}=\sqrt4\cdot\sqrt{-1}=2i$, never $\pm2$.
• Treating $i$ like an ordinary variable and writing $i^2=1$ — by definition $i^2=-1$, always negative.
• Reducing $i^n$ by looking at the exponent's last digit or by guessing, instead of dividing by 4 for the remainder — e.g. $i^{50}=i^{4\times12+2}=i^2=-1$, not $1$.
• Assuming "real" and "complex" are separate, non-overlapping categories — every real number is complex with $b=0$, so $\mathbb{R}\subset\mathbb{C}$, not $\mathbb{R}$ and $\mathbb{C}$ side by side.

Ask the tutor: Still unsure why mathematicians "invent" a number whose square is negative, or want more practice reducing powers of $i$ or splitting real and imaginary parts — ask the AI tutor to walk through a few more examples with you.