Arithmetic of Complex Numbers: Practice

Add the real parts and the imaginary parts separately, keeping each sign attached to its term.

$(7 - 2i) + (1 + 6i) = (7 + 1) + (-2 + 6)\,i = 8 + 4i$

The imaginary coefficients are $-2$ and $6$, and their sum is $4$.

The minus sign flips both terms of the second number, so $-(2 - 7i)$ becomes $-2 + 7i$.

$(6 - 3i) - (2 - 7i) = 6 - 3i - 2 + 7i = (6 - 2) + (-3 + 7)\,i = 4 + 4i$

Forgetting to flip the $-7i$ is what produces the wrong answer $4 - 10i$.

First perform the addition by combining like terms.

$(3 + 2i) + (5 - 9i) = (3 + 5) + (2 - 9)\,i = 8 - 7i$

The result is $8 - 7i$, and its real part is the real number $8$. The imaginary part is $-7$.

Multiplying by the real number $3$ distributes across both terms.

$3(2 + 5i) = 3 \cdot 2 + 3 \cdot 5i = 6 + 15i$

Both the real part and the imaginary part get tripled.

This is the one new rule of complex arithmetic, the defining property of $i$.

$i \cdot i = i^2 = -1$

Every complex multiplication ultimately leans on this single fact.

Multiply the real coefficients and the $i$ factors separately.

$(2i)(5i) = 10\,i^2 = 10(-1) = -10$

The product of two purely imaginary numbers is real, because the $i^2$ becomes $-1$.

Combine like terms; the imaginary parts are opposites.

$(1 + i) + (1 - i) = (1 + 1) + (1 - 1)\,i = 2$

A number plus its conjugate is always real, and equals twice the real part.

Distribute $i$ across both terms, then exchange $i^2$ for $-1$.

$i(3 + 4i) = 3i + 4i^2 = 3i - 4 = -4 + 3i$

The term $4i^2$ leaves the imaginary column and becomes the real number $-4$.

Write the number in the form $a + bi$ first, so $6i = 0 + 6i$.

$\overline{0 + 6i} = 0 - 6i = -6i$

A purely imaginary number's conjugate is its opposite.

A number times its conjugate gives the real number $a^2 + b^2$.

$(3 + 4i)(3 - 4i) = 3^2 + 4^2 = 9 + 16 = 25$

Expanding confirms it: $9 - 12i + 12i - 16i^2 = 9 + 16 = 25$, since $-16i^2 = +16$.

Dividing by a real number divides each part separately.

$\frac{8 + 6i}{2} = \frac{8}{2} + \frac{6}{2}\,i = 4 + 3i$

Both the real part and the imaginary part are halved.

Two complex numbers are equal exactly when their real parts match and their imaginary parts match.

$x = 7, \qquad y = -3$

The imaginary part is the real coefficient of $i$, which here is $-3$, not $-3i$.