Why the Square Root of a Negative Number Is Not Impossible

The equation $x^2+1=0$ has no real solution, since no real number squares to a negative. Introducing $i$ with $i^2=-1$ gives such equations solutions and creates the complex numbers, which are used in alternating current theory, signal processing and quantum physics.

This part covers complex number arithmetic, conjugate, modulus and argument, inverses, and square roots.

**A complex number is written $z=a+ib$ with real part a and imaginary part b; complex numbers are added and subtracted part by part, multiplied like binomials using $i^2=-1$, and divided by multiplying above and below by the conjugate of the denominator.

Powers of i.** $i^2=-1$, $i^3=-i$ and $i^4=1$, so the pattern repeats every four powers: $i^{4k+r}=i^r$.

Worked example. Let $z_1=3+4i$ and $z_2=1-2i$.

- $z_1+z_2=4+2i$
- $z_1-z_2=2+6i$
- $z_1{z}_2=3-6i+4i-8i^2=11-2i$

Division:

$$\frac{3+4i}{1-2i}=\frac{(3+4i)(1+2i)}{(1-2i)(1+2i)}=\frac{3+10i+8i^2}{1+4}=\frac{-5+10i}{5}=-1+2i$$

Check. $(-1+2i)(1-2i)=-1+4i-4i^2=3+4i$.

An everyday example. Engineers at a power substation describe alternating voltages and currents as complex numbers, so that one division gives both the size and the timing shift of the current.

The substance. **$\sqrt{-4}\times \sqrt{-9}$ is not $\sqrt{36}$** — it equals $2i\times 3i=-6$, because $\sqrt{a}\sqrt{b}=\sqrt{ab}$ fails when both numbers are negative.

**The conjugate of $z=a+ib$ is $\bar{z}=a-ib$, its modulus is $|z|=\sqrt{a^2+b^2}$, and its argument is the angle $\theta$ with $\cos \theta =\frac{a}{|z|}$ and $\sin \theta =\frac{b}{|z|}$, taken in $(-\pi ,\pi ]$ as the principal value.

Properties:**

- $z\bar{z}=|z{|}^2$
- $\overline{z_1+z_2}={\bar{z}}_1+{\bar{z}}_2$ and $\overline{z_1{z}_2}={\bar{z}}_1{\bar{z}}_2$
- $|z_1{z}_2|=|z_1||z_2|$ and $\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$
- $|z_1+z_2|\le |z_1|+|z_2|$, the triangle inequality
- $\arg (z_1{z}_2)=\arg z_1+\arg z_2$, adjusted to the principal range

Worked example. For $z=-1+i\sqrt{3}$:

- $\bar{z}=-1-i\sqrt{3}$
- $|z|=\sqrt{1+3}=2$
- $\cos \theta =-\frac{1}{2}$ and $\sin \theta =\frac{\sqrt{3}}{2}$, so z lies in the second quadrant and $\arg z=\frac{2\pi }{3}$

Worked example 2. $|(3+4i)(5-12i)|=5\times 13=65$, without multiplying out.

An everyday example. A delivery rider's trip of 3 km east and 4 km north has modulus 5 km, the straight-line distance, while the argument gives the direction.

The substance. **The argument is not simply ${\tan }^{-1}\frac{b}{a}$** — for $-1+i\sqrt{3}$ that gives $-\frac{\pi }{3}$, which is in the wrong quadrant.

**The additive inverse of $z=a+ib$ is $-z=-a-ib$, since their sum is 0, and the multiplicative inverse of a non-zero z is $z^{-1}=\frac{\bar{z}}{|z{|}^2}=\frac{a-ib}{a^2+b^2}$, since their product is 1.

Why the formula works.**

$$z\times \frac{\bar{z}}{|z{|}^2}=\frac{z\bar{z}}{|z{|}^2}=\frac{|z{|}^2}{|z{|}^2}=1$$

Worked example. Find both inverses of $z=2-3i$.

- Additive inverse: $-2+3i$, since $(2-3i)+(-2+3i)=0$
- $|z{|}^2=4+9=13$ and $\bar{z}=2+3i$
- Multiplicative inverse: $z^{-1}=\frac{2+3i}{13}=\frac{2}{13}+\frac{3}{13}i$

Check. $(2-3i)(2+3i)=4+9=13$, and dividing by 13 gives 1. In particular, the multiplicative inverse of $i$ is $-i$, since $i\times (-i)=-i^2=1$.

An everyday example. Undoing a price change on a shop bill needs the opposite step — subtracting back an added charge, or dividing back a multiplied one — just as inverses undo addition and multiplication.

The substance. 0 has an additive inverse but no multiplicative inverse — the formula needs $|z{|}^2\ne 0$.

**To find $\sqrt{a+ib}$, set it equal to $x+iy$, square to get $x^2-y^2=a$ and $2xy=b$, use $x^2+y^2=\sqrt{a^2+b^2}$ to find x and y, and choose signs so that xy has the same sign as b.

Method:**

- $(x+iy{)}^2=a+ib$ gives $x^2-y^2=a$ and $2xy=b$
- Taking moduli gives $x^2+y^2=|a+ib|$
- Add and subtract to find $x^2$ and $y^2$

Worked example. Find $\sqrt{5+12i}$.

- $x^2-y^2=5$ and $x^2+y^2=\sqrt{25+144}=13$
- Adding, $2x^2=18$, so $x=3$ or $-3$; subtracting, $2y^2=8$, so $y=2$ or $-2$
- $2xy=12$ is positive, so x and y have the same sign

$$\sqrt{5+12i}=\pm (3+2i)$$

Check. $(3+2i{)}^2=9+12i+4i^2=5+12i$.

Worked example 2. For $\sqrt{-8-6i}$: $x^2-y^2=-8$ and $x^2+y^2=10$, so $x^2=1$ and $y^2=9$; since $2xy=-6$ is negative, the roots are $\pm (1-3i)$.

An everyday example. Closing a folding hand fan halfway back undoes a doubled opening — a complex square root similarly halves the argument and takes the square root of the modulus.

The substance. Every non-zero complex number has exactly two square roots, negatives of each other — so the answer always carries $\pm$.