Arithmetic of Complex Numbers: Practice

Use the binomial square pattern, then spend $i^2 = -1$.

$(2 - 3i)^2 = 4 - 12i + 9i^2 = 4 - 12i - 9$

The $9i^2$ becomes $-9$, so the real part is $4 - 9 = -5$.

$(2 - 3i)^2 = -5 - 12i$

This is a product of conjugates, so it equals $a^2 + b^2$ with $a = 5$ and $b = 2$.

$(5 - 2i)(5 + 2i) = 5^2 + 2^2 = 25 + 4 = 29$

The cross terms cancel, and the $i^2$ turns the difference of squares into a sum of squares.

Expand the square and use $i^2 = -1$.

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

The real parts cancel exactly, leaving a purely imaginary result. This little identity is worth remembering.

Distribute the minus sign through $(2 - 3i)$, then collect real and imaginary parts.

$4 - i - 2 + 3i + 1 + 5i = (4 - 2 + 1) + (-1 + 3 + 5)\,i$

The real parts give $3$ and the imaginary coefficients give $7$.

$3 + 7i$

Multiply the numerator and denominator by the conjugate of the denominator, $1 + i$. The denominator becomes $1^2 + 1^2 = 2$.

$\frac{(2 + i)(1 + i)}{2} = \frac{2 + 2i + i + i^2}{2} = \frac{1 + 3i}{2}$

Dividing each part by $2$ gives $\dfrac{1}{2} + \dfrac{3}{2}i$.

The conjugate of $5i = 0 + 5i$ is $-5i$, and $(5i)(-5i) = -25i^2 = 25$.

$\frac{(3 + 4i)(-5i)}{25} = \frac{-15i - 20i^2}{25} = \frac{20 - 15i}{25}$

Dividing each part by $25$ and reducing gives $\dfrac{4}{5} - \dfrac{3}{5}i$.

A number times its conjugate is $a^2 + b^2$, with $a = 4$ and $b = -3$.

$z\,\overline{z} = (4 - 3i)(4 + 3i) = 4^2 + (-3)^2 = 16 + 9 = 25$

The result is real and positive, as the product of conjugates always is for $z \ne 0$.

First compute the product, spending $i^2 = -1$.

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

Then conjugate: $\overline{5 - i} = 5 + i$. You could instead multiply the conjugates first, $(3 - 2i)(1 + i) = 3 + 3i - 2i - 2i^2 = 5 + i$, because conjugation respects products.

Divide both sides by $1 + i$, multiplying the top and bottom by the conjugate $1 - i$; the denominator becomes $1^2 + 1^2 = 2$.

$z = \frac{2i(1 - i)}{2} = i(1 - i) = i - i^2 = 1 + i$

Check: $(1 + i)(1 + i) = (1 + i)^2 = 2i$, as required.

Multiply the numerator and denominator by the conjugate $2 - i$. The denominator becomes $2^2 + 1^2 = 5$.

$\frac{(5 + 5i)(2 - i)}{5} = \frac{10 - 5i + 10i - 5i^2}{5} = \frac{15 + 5i}{5}$

Dividing each part by $5$ gives $3 + i$. Check: $(3 + i)(2 + i) = 6 + 3i + 2i + i^2 = 5 + 5i$.

Compute the reciprocal with the conjugate; the denominator becomes $1^2 + 1^2 = 2$.

$\frac{1}{1 + i} = \frac{1 - i}{2} = \frac{1}{2} - \frac{1}{2}\,i$

The real part of this number is $\dfrac{1}{2}$. The last option is the whole number, not its real part.

Write $z = a + bi$. Then $z + \overline{z} = (a + bi) + (a - bi) = 2a$, which is real no matter what $z$ is.

$z + \overline{z} = 2a$

The others fail for $z = 1 + i$: there $z - \overline{z} = 2i$, $iz = i(1 + i) = -1 + i$, and $z + i = 1 + 2i$, none of which is real.