(2 + 3i) and (1 - i): sum, difference, product and quotient, step by step

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The problem

Work out $z + w$ , $z - w$ , $zw$ and $\dfrac{z}{w}$ .

z = 2 + 3 i, \qquad w = 1 - i

The result

z + w = 3 + 2 i, \quad z - w = 1 + 4 i, \quad zw = 5 + i, \quad \dfrac{z}{w} = -\frac{1}{2} + \frac{5}{2} i

The quotient was checked by multiplying it by $w$ , which gives back $z$ exactly.

checked 27 lines of this solution were evaluated with exact arithmetic before it was shown to you. A solution with a false line is never served.

Picture 1 of 6Step 1: Add: real part with real part, imaginary part with imaginary part

z, w and their sum on the plane. Adding w carries z along the arrow of w: across by the real part of w, then up by its imaginary part.

Add: real part with real part, imaginary part with imaginary part

Why this step

A complex number is two ordinary numbers travelling together: how far across, and how far up. No amount of adding ordinary numbers ever produces a multiple of $i$ , so the two parts cannot leak into each other. That is why addition needs no new rule at all: it is two ordinary additions done side by side.

\displaystyle \left( 2 + 3 i \right) + \left( 1 - i \right)

=

\displaystyle \left( 2 + 1 \right) + \left( 3 - 1 \right) i

Drop the brackets and put like with like: the two real parts together, the two numbers in front of

i

together. checked

=

\displaystyle 3 + 2 i

Two ordinary additions. checked

z, w and their sum on the plane. Adding w carries z along the arrow of w: across by the real part of w, then up by its imaginary part.

Subtract: the minus sign reaches both parts of w

Why this step

Subtracting $w$ means adding its opposite, and the opposite of $c + di$ is $-c - di$ : BOTH parts change sign. The slip to guard against is changing the sign of the real part only and leaving the $i$ part as it was.

\displaystyle \left( 2 + 3 i \right) - \left( 1 - i \right)

=

\displaystyle \left( 2 - 1 \right) + \left( 3 - \left( -1 \right) \right) i

Real part minus real part, and the number in front of

i

minus the number in front of

i

. checked

=

\displaystyle 1 + 4 i

Two ordinary subtractions. checked

Subtracting reverses the arrow of w before carrying z along it, which is why the minus sign has to reach both parts.

Multiply: every part of z meets every part of w

Why this step

The tempting guess is to multiply real by real and imaginary by imaginary, as addition did. It is wrong, and the reason is the distributive law: a bracket times a bracket means each term of the first times each term of the second, which is FOUR products, not two. One of the four is $i$ times $i$ , and that is where the new number earns its keep: it turns back into an ordinary number.

\displaystyle \left( 2 + 3 i \right) \left( 1 - i \right)

=

\displaystyle 2 \cdot 1 + 2 \left( -i \right) + 3 i \cdot 1 + 3 i \left( -i \right)

The distributive law, twice: each of the two terms of

z

multiplies each of the two terms of

w

. Four products. checked

=

\displaystyle 2 - 2 i + 3 i - 3 i^{2}

Work out each of the four products. Numbers multiply as usual, and

i \cdot i

is written

i^2

for one more line so nothing is lost. checked

=

\displaystyle 2 - 2 i + 3 i + 3

Now the one fact that defines

i

i^2 = -1

. The last term stops being imaginary and becomes an ordinary number with its sign flipped. checked

=

\displaystyle \left( 2 + 3 \right) + \left( -2 + 3 \right) i

Collect the ordinary numbers together, and collect the multiples of

i

together. checked

=

\displaystyle 5 + i

Do the two small sums. checked

Multiplying is not carrying at all: the lengths multiply and the directions add, which is why every part of z has to meet every part of w.

Divide: clear i from the bottom with the conjugate

Why this step

Dividing by an ordinary number is easy: divide each part. Dividing by a number with $i$ in it is not, so the plan is to turn the bottom INTO an ordinary number without changing the fraction. The tool is the conjugate of $w$ , written $\overline{w}$ : the same number with the sign of its $i$ part flipped. A number times its conjugate is always $c^2 + d^2$ , an ordinary positive number, because the two middle terms cancel and $i^2 = -1$ repairs the last one. Multiplying top and bottom by the same nonzero number is multiplying by 1, so the value of the fraction does not move.

\frac{2 + 3 i}{1 - i} = \frac{\left( 2 + 3 i \right) \left( 1 + i \right)}{\left( 1 - i \right) \left( 1 + i \right)}

Multiply the top and the bottom by

\overline{w}

. This is multiplying by 1, so nothing has changed except how the fraction looks. checked

\displaystyle \left( 1 - i \right) \left( 1 + i \right)

=

\displaystyle 1 + i - i - i^{2}

The bottom first. Four products again, exactly as in the multiplication step. checked

=

\displaystyle 1 + 1

The two middle terms are equal and opposite, so they cancel. And

i^2 = -1

turns the last term positive. No

i

is left: that was the whole point of the conjugate. checked

=

\displaystyle 2

An ordinary positive number. It is the square of the length of

w

. checked

\displaystyle \left( 2 + 3 i \right) \left( 1 + i \right)

=

\displaystyle 2 \cdot 1 + 2 i + 3 i \cdot 1 + 3 i i

Now the top,

z

times

\overline{w}

: four products, as before. checked

=

\displaystyle 2 + 2 i + 3 i + 3 i^{2}

Work out each of the four products. Numbers multiply as usual, and

i \cdot i

is written

i^2

for one more line so nothing is lost. checked

=

\displaystyle 2 + 2 i + 3 i - 3

Now the one fact that defines

i

i^2 = -1

. The last term stops being imaginary and becomes an ordinary number with its sign flipped. checked

=

\displaystyle \left( 2 - 3 \right) + \left( 2 + 3 \right) i

Collect the ordinary numbers together, and collect the multiples of

i

together. checked

=

\displaystyle -1 + 5 i

Do the two small sums. checked

\displaystyle \frac{-1 + 5 i}{2}

=

\displaystyle \frac{-1}{2} + \frac{5}{2} i

The bottom is now an ordinary number, so it divides each part of the top separately. checked

=

\displaystyle -\frac{1}{2} + \frac{5}{2} i

Simplify the two fractions. checked

The conjugate is the mirror image of w in the real axis: the same distance from the origin, the same real part, the imaginary part flipped. That is why multiplying the two together kills the i: the parts that point up and down cancel exactly, and only a plain positive number is left.

Check the division by multiplying back

Why this step

Division is DEFINED by multiplication: $z / w$ is the number that gives $z$ when it is multiplied by $w$ . So the quotient can be tested with nothing but the multiplication step, and a slip anywhere in the division would show up here.

\displaystyle \left( -\frac{1}{2} + \frac{5}{2} i \right) \left( 1 - i \right)

=

\displaystyle -\frac{1}{2} \cdot 1 + \left( -\frac{1}{2} \left( -i \right) \right) + \frac{5}{2} i \cdot 1 + \frac{5}{2} i \left( -i \right)

The quotient times

w

: four products. checked

=

\displaystyle -\frac{1}{2} + \frac{1}{2} i + \frac{5}{2} i - \frac{5}{2} i^{2}

Work out each of the four products. Numbers multiply as usual, and

i \cdot i

is written

i^2

for one more line so nothing is lost. checked

=

\displaystyle -\frac{1}{2} + \frac{1}{2} i + \frac{5}{2} i + \frac{5}{2}

Now the one fact that defines

i

i^2 = -1

. The last term stops being imaginary and becomes an ordinary number with its sign flipped. checked

=

\displaystyle \left( -\frac{1}{2} + \frac{5}{2} \right) + \left( \frac{1}{2} + \frac{5}{2} \right) i

Collect the ordinary numbers together, and collect the multiples of

i

together. checked

=

\displaystyle 2 + 3 i

Do the two small sums. checked

The check: multiplying the answer back by w must land exactly on z again, which is what makes the division right rather than merely plausible.

Two facts worth seeing on these numbers: the mirror and the ruler

Why this step

The conjugate is a mirror and the modulus is a ruler, and both behave perfectly under multiplication. These two facts are what make the division trick work every time, so it is worth watching them hold on the numbers just used.

\left( 2 - 3 i \right) \left( 1 + i \right) = 5 - i

The mirror of a product is the product of the mirrors: multiply

\overline{z}

\overline{w}

and the answer is the conjugate of

zw

from the multiplication step. checked

13 \cdot 2 = 26

Lengths multiply. The square of the length of

z

13

, of

w

2

, and of

zw

26

. Squares of lengths are compared so that no square root has to be taken. checked

Both numbers with their distances from the origin, which is what the modulus measures and what the two facts in this step are about.

Add, subtract, multiply and divide complex numbers