Arithmetic of Complex Numbers: Additional Practice (Set 2)

Square first: $(2 + i)^2 = 4 + 4i + i^2 = 3 + 4i$.

Then multiply by the remaining factor.

$(3 + 4i)(2 + i) = 6 + 3i + 8i + 4i^2 = 6 + 11i - 4 = 2 + 11i$

This is the very computation Bombelli used in 1572 to rescue the cubic formula.

Multiply both sides by $2 + i$ to undo the division.

$z = (3 - i)(2 + i) = 6 + 3i - 2i - i^2 = 6 + i + 1$

The $-i^2$ becomes $+1$, so $z = 7 + i$.

Divide by $3 - 2i$ using its conjugate $3 + 2i$; the denominator becomes $3^2 + 2^2 = 13$.

$z = \frac{13i(3 + 2i)}{13} = i(3 + 2i) = 3i + 2i^2 = -2 + 3i$

Check: $(3 - 2i)(-2 + 3i) = -6 + 9i + 4i - 6i^2 = -6 + 13i + 6 = 13i$.

Build the cube in stages: $(1 - i)^2 = 1 - 2i + i^2 = -2i$, so

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

The fraction is therefore $\dfrac{2 + 2i}{-2 - 2i} = \dfrac{2 + 2i}{-(2 + 2i)} = -1$.

Substitute $z = 3 + 2i$. First the square:

$(3 + 2i)^2 = 9 + 12i + 4i^2 = 5 + 12i$

Then the whole left side:

$(5 + 12i) - 6(3 + 2i) + 13 = 5 + 12i - 18 - 12i + 13 = 0$

Both parts vanish, so $3 + 2i$ is a solution. (Its conjugate $3 - 2i$ is the other solution, but it is not offered.)

Divide both sides by $3 + i$ using the conjugate $3 - i$; the denominator becomes $3^2 + 1^2 = 10$.

$x + yi = \frac{10i(3 - i)}{10} = i(3 - i) = 3i - i^2 = 1 + 3i$

Matching parts gives $x = 1$ and $y = 3$. Check: $(1 + 3i)(3 + i) = 3 + i + 9i + 3i^2 = 10i$.

Conjugation is its own undo: conjugating twice returns the original number.

$z = \overline{\overline{z}} = \overline{4 + 9i} = 4 - 9i$

Only the sign of the imaginary part flips; the real part stays $4$.

Multiply first, following the order of operations.

$(4 - 5i)(1 + i) = 4 + 4i - 5i - 5i^2 = 4 - i + 5 = 9 - i$

Then subtract, distributing the minus sign through both parts.

$(6 - 2i) - (9 - i) = (6 - 9) + (-2 + 1)\,i = -3 - i$

Write $z = a + bi$. Then $z + \overline{z} = 2a = 10$, so $a = 5$, and $z\,\overline{z} = a^2 + b^2 = 29$.

$25 + b^2 = 29 \ \Rightarrow \ b^2 = 4 \ \Rightarrow \ b = 2 \ \text{or} \ b = -2$

So $z$ is $5 + 2i$ or $5 - 2i$, and only $5 + 2i$ is offered. The choice $5 + 4i$ has the right sum but $z\,\overline{z} = 41$.

First square: $(2 + i)^2 = 4 + 4i + i^2 = 3 + 4i$. Then conjugate.

$\overline{3 + 4i} = 3 - 4i$

Because conjugation respects products, you could instead square the conjugate first: $(2 - i)^2 = 4 - 4i + i^2 = 3 - 4i$, the same answer.

Substitute $z = -1 + 2i$. The square is

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

Then the whole left side collapses:

$(-3 - 4i) + 2(-1 + 2i) + 5 = (-3 - 2 + 5) + (-4 + 4)\,i = 0$

So $-1 + 2i$ is a solution. (Its conjugate $-1 - 2i$ is the other one, but it is not offered.)

Square the candidate $5 - 2i$:

$(5 - 2i)^2 = 25 - 20i + 4i^2 = 25 - 20i - 4 = 21 - 20i$

It works. The others fail: $(5 + 2i)^2 = 21 + 20i$, while $(2 - 5i)^2$ and $(-2 + 5i)^2$ both equal $-21 - 20i$.