Arithmetic with Complex Numbers: Practice

A complex number in standard form is $a + bi$, and its imaginary part is the real coefficient $b$, not $bi$. Write $7 - 4i = 7 + (-4)i$.

$\operatorname{Im}(7 - 4i) = -4$

The real part is $7$; the imaginary part is the number $-4$.

In standard form $a + bi$, the real part is $a$. Here $a = -3$ and $b = 8$.

$\operatorname{Re}(-3 + 8i) = -3$

The $8$ is the imaginary part, not the real part.

Combine the real parts and the imaginary parts separately.

$(2 + 3i) + (5 + i) = (2 + 5) + (3 + 1)i = 7 + 4i$

The imaginary part of the second number is $1$, so the imaginary parts add to $3 + 1 = 4$.

Distribute the $i$, then replace $i^2$ with $-1$.

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

The $2i^2$ term becomes $-2$, which is the new real part.

Expand term by term, then replace $i^2$ with $-1$.

$(1 + i)(2 + i) = 2 + i + 2i + i^2 = 2 + 3i - 1 = 1 + 3i$

The $i^2$ becomes $-1$, lowering the real part from $2$ to $1$.

The conjugate flips the sign of the imaginary part only, leaving the real part unchanged.

$\overline{6 + 2i} = 6 - 2i$

The real part stays $6$; only the $+2i$ becomes $-2i$.

Multiply the coefficients and the $i$ factors, then replace $i^2$ with $-1$.

$(4i)(3i) = 12i^2 = 12(-1) = -12$

The two imaginary factors produce $i^2 = -1$, so the result is the real number $-12$.

Distribute the $2i$, then replace $i^2$ with $-1$.

$2i(3 - 4i) = 6i - 8i^2 = 6i + 8 = 8 + 6i$

The term $-8i^2$ becomes $+8$; leaving it as $-8$ forgets that $i^2 = -1$.

Write the pure imaginary number in standard form: $9i = 0 + 9i$.

$\operatorname{Re}(9i) = 0$

A pure imaginary number has real part $0$; the $9$ is its imaginary part.

Expand term by term, then replace $i^2$ with $-1$.

$(5 - 2i)(1 + i) = 5 + 5i - 2i - 2i^2 = 5 + 3i + 2 = 7 + 3i$

The term $-2i^2$ becomes $+2$, raising the real part from $5$ to $7$.

A number times its conjugate clears the imaginary part.

$(2 + i)(2 - i) = 2^2 + 1^2 = 5$

The others stay complex: $(2 + i) + (2 + i) = 4 + 2i$, and $i(1 + i) = -1 + i$, and $3 + i$ already has an imaginary part.

Distribute the $i$, then replace $i^2$ with $-1$.

$i(2 - i) = 2i - i^2 = 2i + 1 = 1 + 2i$

The term $-i^2$ becomes $+1$, which is the real part of the answer.