Pj4, Pj5 - Complex arithmetic and equating parts

Adding, subtracting and multiplying

Which complex numbers square to give $i$?

You can already find the square roots of any negative number: $-9$ has the two square roots $3i$ and $-3i$. But $i$ is not a negative number, or even a real one, so that rule does not help. To test a guess you must be able to multiply complex numbers, which is the job of this section. To find the answer without guessing you need one more idea, which comes in the last section.

You add, subtract and multiply complex numbers exactly as you handle brackets in algebra, treating $i$ like a letter and then replacing $i^2$ by $-1$.

Adding and subtracting combine the real parts and the imaginary parts separately: $(2 + 5i) + (1 - 3i) = 3 + 2i$ and $(2 + 5i) - (1 - 3i) = 1 + 8i$. Multiplying uses all four products from the two brackets, just as $(a + b)(c + d) = ac + ad + bc + bd$. The one new step is that the term in $i^2$ becomes real: $bi \times di = bdi^2 = -bd$.

A grid keeps track of the four products. Here it is for $(2 + 3i)(4 - i)$:

[DIAGRAM: asset_name: Multiplying two complex numbers with a grid; asset_slug: complex_product_grid; caption: A grid shows the four products when two complex numbers are multiplied, and the i squared term turning real; description: A two by two grid multiplies 2 plus 3i, down the side, by 4 minus i, along the top. The cells hold 8, minus 2i, 12i and minus 3i squared, which equals plus 3. The cells 8 and plus 3 are shaded blue as real terms, the cells minus 2i and 12i are shaded lavender as imaginary terms, and the plus 3 cell has an ochre border with the note that i squared equals minus 1 turns this term real. A side panel adds the real terms, 8 plus 3 equals 11, and the imaginary terms, minus 2i plus 12i equals 10i, giving the product 11 plus 10i.]

The same expansion tests a first guess at a square root of $i$: $(1 + i)^2 = 1 + i + i + i^2 = 2i$. That is twice $i$, so $1 + i$ is close but not right.

Worked example 2 changes one feature: the number is raised to a power. A power is repeated multiplication of the whole number, so you multiply in stages.

This exact calculation has a famous history. Cardano's formula for the cubic equation $x^3 = 15x + 4$ produces $\sqrt[3]{2 + 11i} + \sqrt[3]{2 - 11i}$, even though the equation has the obvious root $x = 4$. In his book L'Algebra (1572), Rafael Bombelli saw that $(2 + i)^3 = 2 + 11i$ and $(2 - i)^3 = 2 - 11i$, so the expression is just $(2 + i) + (2 - i) = 4$. A route through imaginary numbers had led to a real answer.

Subtraction causes the other frequent error: the minus sign applies to both parts of the second number, so keep that number in brackets until you have removed them.

Dividing complex numbers

To divide by a complex number, multiply the numerator and the denominator by the conjugate of the denominator, so that the denominator becomes a real number.

Multiplying the top and the bottom of a fraction by $w^*$, the conjugate of the denominator, does not change the value of the fraction, because you are multiplying by $\frac{w^*}{w^*} = 1$. It does turn the denominator into the real number $ww^*$, and dividing by a real number is easy: divide each part.

A quick case first. To work out $\frac{10}{1 + 3i}$, multiply the top and bottom by $1 - 3i$. The denominator becomes $(1 + 3i)(1 - 3i) = 1 + 9 = 10$, so $\frac{10}{1 + 3i} = \frac{10(1 - 3i)}{10} = 1 - 3i$.

Worked example 4 changes one feature: surds appear in the imaginary parts, and the question carries the instruction to show detailed reasoning. Take care when squaring a surd times $i$: $(i\sqrt{2})^2 = i^2 \times 2 = -2$.

The instruction “In this question you must show detailed reasoning” changes what earns marks, so it deserves its own habit.

Equating real and imaginary parts

A complex number is zero only when its real part and its imaginary part are both zero, so one equation between complex numbers gives two equations between real numbers.

Start with a concrete case. Suppose $x$ and $y$ are real and $(x - 3) + (y + 2)i = 0$. A real part cannot cancel an imaginary part, so both brackets must be zero: $x = 3$ and $y = -2$.

Why can a real part never cancel an imaginary part? Suppose $a + bi = 0$, where $a$ and $b$ are real and $b \ne 0$. Then $i = -\frac{a}{b}$, which is a real number. The square of a real number is never negative, but $i^2 = -1$. That is impossible, so $b = 0$, and then $a = a + 0i = 0$ as well. On an Argand diagram the same fact is easy to see: the number $a + bi$ is the point $(a, b)$, and only the origin is zero.

[DIAGRAM: asset_name: Only the origin is zero; asset_slug: zero_only_at_origin; caption: A complex number is zero only at the origin, where both parts are 0; description: An Argand diagram with axes labelled Re and Im. The origin is marked in green and labelled 0 plus 0i equals 0. Blue points mark 3 plus 0i on the real axis, 0 plus 2i on the imaginary axis and 3 plus 2i. A side panel says that x plus yi is the point (x, y), that 3 plus 0i has imaginary part 0 but a real part that is not 0, that 0 plus 2i has real part 0 but an imaginary part that is not 0, and that only the origin is zero: x plus yi equals 0 exactly when x equals 0 and y equals 0.]

Using this is called equating real and imaginary parts. Write each side in the form $a + bi$, with $a$ and $b$ real, then write one equation for the real parts and one for the imaginary parts.

Worked example 6 changes one feature: the unknown is the complex number $z$ itself, and its conjugate $z^*$ appears too. Write $z = x + yi$ first, so that the unknowns $x$ and $y$ are real and you can equate parts.

Questions like Worked example 6 usually say “Determine”, which means every step has to be visible.

Back to the square root of i

Now the opening question has an answer. Suppose $z = x + yi$, with $x$ and $y$ real, and $z^2 = i$.

The number $i$ is $0 + 1i$, so equating parts gives

The first equation gives $y = x$ or $y = -x$. If $y = -x$, the second becomes $-2x^2 = 1$, which has no solution because $x$ is real. So $y = x$, and $2x^2 = 1$ gives $x = \pm\frac{1}{\sqrt{2}} = \pm\frac{\sqrt{2}}{2}$.

The square roots of $i$ are $\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i$ and $-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i$. Check the first: $\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i\right)^2 = \frac{1}{2} + i + \frac{1}{2}i^2 = i$. The first guess, $1 + i$, squared to $2i$; dividing it by $\sqrt{2}$ halves its square, which is exactly what this answer does.

Quick Recap

• Add and subtract by combining the real parts and the imaginary parts separately; multiply by expanding all four products and replacing $i^2$ by $-1$.
• Raise to a power in stages, multiplying the whole number each time.
• To divide, multiply the top and the bottom by the conjugate of the denominator, which makes the denominator the real number $ww^*$.
• For real $a$ and $b$, $a + bi = 0$ only when $a = 0$ and $b = 0$, so equal complex numbers have equal real parts and equal imaginary parts.
• For an equation in $z$ and $z^*$, write $z = x + yi$ and $z^* = x - yi$, then equate real and imaginary parts.

The square roots of $i$ are $\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i$ and $-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i$.