Dividing, and the powers of i

What you will be able to do

• find the multiplicative inverse of a complex number;
• divide one complex number by another;
• find any power of $i$, positive or negative;
• avoid the trap $\sqrt{a}\sqrt{b} = \sqrt{ab}$ when $a$ and $b$ are both negative.

The multiplicative inverse

The multiplicative inverse of a non-zero number $z$ is the number $\frac{1}{z}$ that gives $z \times \frac{1}{z} = 1$. For $z = a + ib$, the trick is to multiply the top and the bottom by $a - ib$. The bottom becomes real:

So

The number $a - ib$, with the sign of $i$ changed, has a name: the conjugate. You will meet it properly in the next course.

Example. $\dfrac{1}{3 + 4i} = \dfrac{3 - 4i}{9 + 16} = \dfrac{3}{25} - \dfrac{4}{25}i$.

The inverse exists for every complex number except $0$, because $a^2 + b^2 = 0$ only when $a = b = 0$.

Dividing

To divide $z_1$ by $z_2$, multiply $z_1$ by the inverse of $z_2$. In practice, multiply the top and the bottom by $z_2$ with the sign of $i$ changed.

Example. Find $\dfrac{4 + 7i}{1 - 2i}$.

• Multiply the top and the bottom by $1 + 2i$.
• The bottom: $(1 - 2i)(1 + 2i) = 1 + 4 = 5$.
• The top: $(4 + 7i)(1 + 2i) = 4 + 8i + 7i + 14i^2 = -10 + 15i$.
• So the quotient is $\dfrac{-10 + 15i}{5} = -2 + 3i$.

Check by multiplying back: $(-2 + 3i)(1 - 2i) = -2 + 4i + 3i - 6i^2 = 4 + 7i$.

The powers of i

Multiply by $i$ again and again:

• $i^1 = i$
• $i^2 = -1$
• $i^3 = i^2 \times i = -i$
• $i^4 = (i^2)^2 = 1$

After four steps we are back at $1$, and the cycle starts again: $i^5 = i$, $i^6 = -1$, and so on. So to find $i^n$, divide $n$ by $4$ and keep the remainder $r$. Then $i^n = i^r$.

Example. $43 = 4 \times 10 + 3$, so $i^{43} = i^3 = -i$.

Negative powers follow the same cycle, because $i^{-1} = \frac{1}{i} = \frac{-i}{1} = -i$. For example, $i^{-6} = \frac{1}{i^6} = \frac{1}{-1} = -1$. Python's % gives the right remainder here too: -6 % 4 is 2, and $i^2 = -1$.

A square root trap

For positive numbers, $\sqrt{a}\sqrt{b} = \sqrt{ab}$. For two negative numbers it fails:

So always write each square root of a negative number as a multiple of $i$ first, and multiply afterwards.

A worked example

Find $i + i^2 + i^3 + \dots + i^{30}$.

1. Group in fours. Each group of four powers adds up to $i - 1 - i + 1 = 0$.
2. Count the groups. $30 = 4 \times 7 + 2$, so the first $28$ powers form $7$ groups, which add up to $0$.
3. Add what is left. $i^{29} + i^{30} = i^1 + i^2 = i - 1$.

The sum is $-1 + i$.

Practise with new numbers

Try it in Python

The program builds the powers of $i$ one at a time, by multiplying by 1j, and prints the remainder n % 4 next to each. Then it divides two complex numbers with /.

Common mistakes

• Dividing the parts one by one. $\frac{6 + 8i}{2 + 4i}$ is not $3 + 2i$. Multiply the top and the bottom by $2 - 4i$ instead: the answer is $2 - 0.4i$.
• Using the wrong remainder. For $i^{38}$, the remainder of $38$ on division by $4$ is $2$, so $i^{38} = -1$. It is not $i^{8} = 1$ from the last digit.
• Multiplying square roots of negatives under one sign. $\sqrt{-2}\sqrt{-8} = -4$, not $4$.
• Dividing by zero. $\frac{1}{0 + 0i}$ has no meaning. In Python, 1 / 0j stops with ZeroDivisionError.

Remember these

• $\dfrac{1}{a + ib} = \dfrac{a - ib}{a^2 + b^2}$, for $a + ib \neq 0$.
• To divide, multiply the top and the bottom by the bottom with the sign of $i$ changed.
• $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, and then the cycle repeats.
• $\sqrt{a}\sqrt{b} = \sqrt{ab}$ only when $a$ and $b$ are not both negative.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the multiplicative inverse of $2 - i$.

2. Divide $\dfrac{3 + 2i}{2 + 3i}$.

3. Find $i^{97}$ and $i^{-3}$.

4. Ravi writes $\sqrt{-3} \times \sqrt{-12} = \sqrt{36} = 6$. What is the right value?

5. Why does $0$ have no multiplicative inverse?