Pj1, Pj2 - Complex number terms and quadratic roots

Square roots of negative numbers

Can you split 10 into two numbers whose product is 40?

Try a few pairs. 2 and 8 give 16, 4 and 6 give 24, and 5 and 5 give 25. Moving the two numbers apart only makes the product smaller, so with real numbers 25 is the best you can do. The algebra agrees. If the numbers are $x$ and $10 - x$, then

No real number squares to $-15$, so there is no real solution. Gerolamo Cardano posed exactly this problem in his book Ars Magna in 1545. He wrote the answers as $5 + \sqrt{-15}$ and $5 - \sqrt{-15}$, checked that their product really is 40, and then dismissed the calculation as clever but useless. By the end of this lesson you will see that these are two perfectly good numbers, and you will find them in a few lines.

The first step is to give a square root of $-1$ a name.

Check the first example: $(5i)^2 = 25i^2 = 25 \times (-1) = -25$. Also $(-5i)^2 = 25i^2 = -25$, so $-25$ has two square roots, $5i$ and $-5i$, just as $25$ has $5$ and $-5$. The symbol $\sqrt{-25}$ means $5i$, and the two square roots together are $\pm 5i$. With $i$ available, the equation $(x - 5)^2 = -15$ is no longer a dead end; the last section of this lesson finishes Cardano's problem.

Powers of i

Because $i^2 = -1$, every power of $i$ simplifies:

Once you reach $i^4 = 1$, the pattern starts again.

[DIAGRAM: asset_name: Powers of i; asset_slug: i_power_cycle; caption: Multiplying by i four times brings you back to 1; description: Four boxes in a loop show 1 on the right, i at the top, minus 1 on the left and minus i at the bottom, each labelled with the power of i it equals. Anticlockwise arrows marked times i join them, so four steps return to 1 and a power of i depends only on the remainder when the index is divided by 4.]

For example, $23 = 4 \times 5 + 3$, so $i^{23} = (i^4)^5 \times i^3 = 1 \times (-i) = -i$.

The language of complex numbers

A complex number is a real number plus a real multiple of $i$; this section names its parts and pictures it as a point.

Cardano's numbers already have this shape: $5 + i\sqrt{15}$ is the real number 5 plus $\sqrt{15}$ lots of $i$.

The fourth row matters. A real number such as $-8$ is a complex number whose imaginary part is 0, so every real number is also a complex number. A number such as $5i$, whose real part is 0, is called purely imaginary. The set of all complex numbers is written $\mathbb{C}$, and it contains the set of real numbers $\mathbb{R}$.

[DIAGRAM: asset_name: Real numbers inside the complex numbers; asset_slug: complex_number_sets; caption: Every real number is a complex number with imaginary part 0; description: A large lavender region labelled C, all complex numbers x plus yi with x and y real, contains a smaller green region labelled R, real numbers with imaginary part 0, holding minus 8, three quarters, root 2 and pi. Outside R but inside C are 4 minus 7i, 5i, 5 plus i root 15 and minus 1 minus i, each with its real and imaginary parts written beneath it.]

You can read off the parts only when the number is written as one real number plus one real multiple of $i$. For example, $2i(3 - i)$ is not yet in that form:

so its real part is 2 and its imaginary part is 6.

Two points of notation. Exam papers write the same number as $x + yi$, $x + iy$ or $a + bi$; all three mean the same thing. When the imaginary part is a surd, write $i\sqrt{15}$ or $\sqrt{15}\,i$, never $\sqrt{15i}$, so that the $i$ is clearly outside the root.

Complex numbers as points

Because $z = x + yi$ is fixed by the two real numbers $x$ and $y$, you can plot it as the point $(x, y)$. The horizontal axis is the real axis, labelled Re, and the vertical axis is the imaginary axis, labelled Im. This picture is called an Argand diagram, after Jean-Robert Argand, who published it in 1806. Real numbers lie on the real axis and purely imaginary numbers lie on the imaginary axis.

[DIAGRAM: asset_name: A complex number as a point; asset_slug: argand_modulus_argument; caption: The complex number 3 plus 4i drawn as a point, with its modulus and argument; description: An Argand diagram with the horizontal real axis labelled Re and the vertical imaginary axis labelled Im. The point z equals 3 plus 4i is marked, with dashed lines showing Re(z) equals 3 across and Im(z) equals 4 up. A blue line from the origin O to z is labelled modulus of z equals 5, and a green arc at O from the positive real axis to that line is labelled arg z. A side panel says that the modulus is the distance from O, the square root of 3 squared plus 4 squared, and that the argument is the angle from the positive real axis, with tan of arg z equal to four thirds, so arg z is about 0.927 radians, and that arguments lie between minus pi and pi.]

Two more words describe where the point is: how far it is from the origin $O$, and in which direction.

The modulus is Pythagoras' theorem on the right-angled triangle in the diagram, so it is always real and never negative. For a real number it is the absolute value you already know: $|-7| = 7$.

For $z = 3 + 4i$, which lies in the first quadrant, $\tan(\arg z) = \frac{4}{3}$, so $\arg z = \arctan\frac{4}{3} \approx 0.927$. In the other quadrants this shortcut needs care, and the number 0 has no argument at all. The lesson on modulus-argument form gives the full method.

The complex conjugate

Every complex number has a partner, its conjugate, with the same real part and the opposite imaginary part, and together the pair has a real sum and a real product.

Start with $z = 3 + 2i$ and change the sign of its imaginary part to get $3 - 2i$. Then

Both answers are real. In the sum the imaginary parts cancel. In the product the two cross terms cancel, and $-4i^2$ becomes $+4$.

On an Argand diagram, $z^*$ is the reflection of $z$ in the real axis, so $z$ and $z^*$ are the same distance from the origin: $|z^*| = |z|$.

The same cancelling happens for every complex number. With $z = x + yi$:

The number $x^2 + y^2$ is a sum of squares of two real numbers, so $zz^*$ is real and never negative, and it is zero only when $z = 0$. It is also the square of the modulus: $zz^* = |z|^2$, the squared distance from the origin to $z$.

This is the complex version of a trick you already use with surds: $2 + \sqrt{3}$ and $2 - \sqrt{3}$ multiply to the rational number 1. In the next lesson the same trick is the key to dividing by a complex number.

Quadratic equations with real coefficients

Every quadratic equation with real coefficients has two roots among the complex numbers, and when the discriminant is negative they form a conjugate pair $p + qi$ and $p - qi$.

With $i$, a negative discriminant is no longer a dead end. If $b^2 - 4ac < 0$, then $4ac - b^2$ is positive and $\sqrt{b^2 - 4ac} = i\sqrt{4ac - b^2}$. The $\pm$ in the formula then produces two roots with the same real part and opposite imaginary parts.

To check a root, substitute it. You need only two moves: square the root by expanding the bracket with $i^2 = -1$, then collect the real terms and the terms in $i$.

The graph shows where these roots come from. Completing the square gives $x^2 + 4x + 13 = (x + 2)^2 + 9$, so the parabola has its vertex at $(-2, 9)$ and never meets the $x$-axis. The line of symmetry $x = -2$ gives the real part of each root, and the imaginary parts, $\pm 3$, are the square roots of the vertex height 9. This picture works whenever the coefficient of $x^2$ is 1.

[DIAGRAM: asset_name: A parabola with complex roots; asset_slug: parabola_complex_roots; caption: A parabola that never meets the x-axis still has two complex roots; description: Graph of y equals x squared plus 4x plus 13 with vertex at (minus 2, 9) above the x-axis. A dashed vertical line x equals minus 2 marks the axis of symmetry and a double arrow marks the vertex height 9. A side panel explains that the axis of symmetry gives the real part minus 2 and that 9 equals 3 squared gives the imaginary parts plus or minus 3, so the roots are minus 2 plus or minus 3i.]

Worked example 2 changes one feature: the coefficient of $z^2$ is not 1. The formula is quicker than completing the square here, and both terms on the top must be divided by $2a$.

Exam questions also decide how much working you must show, and a calculator can solve quadratics in seconds, so the method has to be visible.

Back to Cardano

Now the opening question has an answer. Two numbers $\alpha$ and $\beta$ are the roots of $(z - \alpha)(z - \beta) = 0$, which expands to $z^2 - (\alpha + \beta)z + \alpha\beta = 0$. With sum 10 and product 40 this is $z^2 - 10z + 40 = 0$, the equation from the first section. Its discriminant is $100 - 160 = -60$, so

The conjugate pair results from earlier confirm the answer:

Cardano's “impossible” numbers are ordinary complex numbers. They cannot be found among the real numbers, but they behave exactly as the problem requires.

Quick Recap

• $i^2 = -1$, so $\sqrt{-k} = i\sqrt{k}$ for $k > 0$, and the powers of $i$ repeat every four.
• A complex number is $z = x + yi$ with $x$ and $y$ real; $\operatorname{Re}(z) = x$ and $\operatorname{Im}(z) = y$ are both real numbers.
• On an Argand diagram $z$ is the point $(x, y)$: $|z| = \sqrt{x^2 + y^2}$ is its distance from the origin and $\arg z$ is its angle from the positive real axis, with $-\pi < \arg z \le \pi$.
• The conjugate $z^* = x - yi$ gives $z + z^* = 2x$ and $zz^* = x^2 + y^2 = |z|^2$.
• A quadratic with real coefficients and a negative discriminant has a conjugate pair of roots $-\frac{b}{2a} \pm \frac{\sqrt{4ac - b^2}}{2a}i$.

Cardano's two numbers with sum 10 and product 40 are $5 + i\sqrt{15}$ and $5 - i\sqrt{15}$.