Arithmetic of Complex Numbers: Additional Practice (Set 1)

Distribute the minus sign through $(1 - i)$, then collect real and imaginary parts.

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

The real parts give $2$ and the imaginary coefficients give $5$.

$2 + 5i$

Each factor pair is a conjugate product, so each equals $a^2 + b^2$.

$(4 - 3i)(4 + 3i) = 16 + 9 = 25, \qquad (2 + i)(2 - i) = 4 + 1 = 5$

Subtracting the two real results gives $25 - 5 = 20$.

Expand the binomial square, then exchange $i^2$ for $-1$.

$(2 - 5i)^2 = 4 - 20i + 25i^2 = 4 - 20i - 25$

The $25i^2$ becomes $-25$, so the real part is $4 - 25 = -21$.

$(2 - 5i)^2 = -21 - 20i$

Multiply the numerator and denominator by $-i$, the conjugate of $i$; note $i(-i) = -i^2 = 1$.

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

Check: $i(1 - 3i) = i - 3i^2 = 3 + i$, as required.

Square twice. First, $(1 - i)^2 = 1 - 2i + i^2 = -2i$.

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

The same answer as $(1 + i)^4$, since the two bases are conjugates and conjugation respects powers.

Rationalize each reciprocal with the conjugate of its denominator; both denominators become $1^2 + 1^2 = 2$.

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

Adding them, the imaginary parts cancel: $\dfrac{(1 - i) + (1 + i)}{2} = \dfrac{2}{2} = 1$.

Convert each square root of a negative to $i$-form first: since $(3i)^2 = 9i^2 = -9$ and $(2i)^2 = 4i^2 = -4$, the roots are $3i$ and $2i$.

$\sqrt{-9} \cdot \sqrt{-4} = (3i)(2i) = 6i^2 = -6$

The shortcut $\sqrt{-9}\cdot\sqrt{-4} = \sqrt{36} = 6$ is wrong, because the rule $\sqrt{x}\,\sqrt{y} = \sqrt{xy}$ is only valid for non-negative $x$ and $y$.

First find the reciprocal with the conjugate: $\dfrac{1}{1 + i} = \dfrac{1 - i}{2} = \dfrac{1}{2} - \dfrac{1}{2}i$.

Then add it to $z$.

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

Multiply the conjugate pair first, because its product is real.

$(3 + 2i)(3 - 2i) = 3^2 + 2^2 = 13$

Then scale the remaining factor.

$13(1 + i) = 13 + 13i$

First, $zw = (2 + 3i)(1 - i) = 2 - 2i + 3i - 3i^2 = 5 + i$.

Because conjugation respects products, $\overline{z}\,\overline{w} = \overline{zw} = 5 - i$; computing directly, $(2 - 3i)(1 + i) = 2 + 2i - 3i - 3i^2 = 5 - i$, which agrees.

$(5 + i) + (5 - i) = 10$

A number plus its conjugate is twice its real part, so the answer had to be real.

Multiply the numerator and denominator by the conjugate $2 + 3i$. The denominator becomes $2^2 + 3^2 = 13$.

$\frac{(8 + i)(2 + 3i)}{13} = \frac{16 + 24i + 2i + 3i^2}{13} = \frac{13 + 26i}{13} = 1 + 2i$

Check: $(1 + 2i)(2 - 3i) = 2 - 3i + 4i - 6i^2 = 8 + i$.

Reduce the first four powers with $i^2 = -1$: they are $i, -1, -i, 1$, and their sum is

$i - 1 - i + 1 = 0$

The next four repeat the same cycle, since $i^5 = i^4 \cdot i = i$ and so on, so they also sum to $0$. The total is $0 + 0 = 0$.