Complex Numbers: Chapter Test

Put the number into standard form $a + bi$ first, so that the coefficient of $i$ is visible with its sign attached.

$8 - i\sqrt{11} = 8 + \left(-\sqrt{11}\right)i$

The imaginary part is the real number $b = -\sqrt{11}$.

The minus sign reaches both parts of the second number, so $-(-2)$ contributes $+2$ and $-(+5i)$ contributes $-5i$.

$(6 - 7i) - (-2 + 5i) = (6 + 2) + (-7 - 5)i = 8 - 12i$

Isolate the square, then take both square roots of a negative number using $\sqrt{-k} = i\sqrt{k}$.

$x^2 = -\frac{81}{4} \quad\Longrightarrow\quad x = \pm\, i\sqrt{\frac{81}{4}} = \pm\frac{9}{2}i$

Both values check: $\left(\tfrac{9}{2}i\right)^2 = \tfrac{81}{4}i^2 = -\tfrac{81}{4}$.

Square each part, add, then take the non-negative square root, in that order. Squaring erases the sign of the real part.

$|-11 + 4i| = \sqrt{(-11)^2 + 4^2} = \sqrt{121 + 16} = \sqrt{137}$

Nothing divides $137$ into a square, so the exact modulus is the surd itself.

A power of $i$ depends only on the remainder of its exponent on division by $4$. Since $62 = 4 \times 15 + 2$ and $79 = 4 \times 19 + 3$:

$i^{62} + i^{79} = i^2 + i^3 = -1 + (-i) = -1 - i$

Expand all four products with their signs, then exchange $i^2$ for $-1$, which sends $-15i^2$ into the real part as $+15$.

$(6 + 5i)(2 - 3i) = 12 - 18i + 10i - 15i^2 = (12 + 15) + (-18 + 10)i = 27 - 8i$

Adding $w$ sends $z$ to $z + w$, so $w$ is the image minus the starting number.

$w = (-1 + 2i) - (5 + 7i) = (-1 - 5) + (2 - 7)i = -6 - 5i$

The slide moves every point of the plane $6$ units left and $5$ units down.

Rewrite each root as $i$ times a real root before multiplying anything, since $\sqrt{a}\sqrt{b} = \sqrt{ab}$ was proved for non-negative radicands only. The two factors of $i$ then give $i^2 = -1$.

$\sqrt{-20} \cdot \sqrt{-5} = \left(2i\sqrt{5}\right)\left(i\sqrt{5}\right) = 2i^2 \cdot 5 = -10$

With $a = 2$, $b = 6$, $c = 5$ the discriminant is $36 - 40 = -4$, so $\sqrt{-4} = 2i$ and the whole numerator is divided by $2a = 4$.

$x = \frac{-6 \pm \sqrt{-4}}{4} = \frac{-6 \pm 2i}{4} = -\frac{3}{2} \pm \frac{1}{2}i$

The shared real part $-\tfrac{3}{2}$ is the axis of symmetry $-\tfrac{b}{2a}$.

Multiply numerator and denominator by the conjugate of the denominator, which is multiplying by $1$. The new denominator is the real number $2^2 + 1^2 = 5$.

$\frac{16 + 3i}{2 + i} = \frac{(16 + 3i)(2 - i)}{5} = \frac{32 - 16i + 6i + 3}{5} = \frac{35 - 10i}{5} = 7 - 2i$

Check by multiplying back: $(7 - 2i)(2 + i) = 14 + 7i - 4i + 2 = 16 + 3i$.

The conjugate is $\overline{z} = 10 - 3i$, and the cross terms cancel while $-b^2i^2$ becomes $+b^2$.

$z\overline{z} = (10 + 3i)(10 - 3i) = 10^2 + 3^2 = 100 + 9 = 109$

This is $|z|^2$, the squared distance from $0$, never negative and real for every $z$.

Real coefficients force the other root to be the conjugate $-8 - 3i$. The sum is $-16$ and the product is the squared modulus.

$(-8 + 3i)(-8 - 3i) = 64 + 9 = 73, \qquad x^2 - (\text{sum})x + (\text{product}) = x^2 + 16x + 73$

Over the complex numbers a quadratic always has two roots counted with multiplicity, so the discriminant no longer settles how many roots there are. What it reports now is position: a negative value sends the pair off the real axis, symmetrically about it, while its vanishing is what makes the two coincide.

$b^2 - 4ac < 0 \quad\Longrightarrow\quad x = -\frac{b}{2a} \pm \frac{\sqrt{4ac - b^2}}{2a}\,i$

The shared real part is the axis of symmetry, and the opposite imaginary parts place one root above the axis and one below.

In Quadrant IV the real part is positive and the imaginary part is negative, so write $z = a + bi$ with $a > 0$ and $b < 0$. Conjugating flips the sign of $b$ only, and the minus sign then flips both parts.

$-\overline{z} = -(a - bi) = -a + bi$

Now $-a < 0$ and $b < 0$, which is Quadrant III.

Expand the left side, spending $i^2 = -1$, and then match parts, which is legitimate because $x$ and $y$ are real.

$(x + yi)(1 + i) = (x - y) + (x + y)i = 5 - i$

So $x - y = 5$ and $x + y = -1$, giving $x = 2$ and $y = -3$. The check is one multiplication: $(2 - 3i)(1 + i) = 2 + 2i - 3i + 3 = 5 - i$.

The modulus of a product is the product of the moduli, so no expansion is needed.

$|2 - 6i| = \sqrt{40} = 2\sqrt{10}, \qquad |9 + 3i| = \sqrt{90} = 3\sqrt{10}, \qquad |zw| = 2\sqrt{10} \cdot 3\sqrt{10} = 60$

The slow route agrees: $(2 - 6i)(9 + 3i) = 36 - 48i$, and $\sqrt{36^2 + 48^2} = \sqrt{3600} = 60$.

The triangle inequality bounds the sum above, and its reverse reading bounds it below, so the length of the sum is confined to a window.

$\bigl||z| - |w|\bigr| \le |z + w| \le |z| + |w| \quad\Longrightarrow\quad 4 \le |z + w| \le 16$

Both ends are attained, by $z = 6$ with $w = -10$ and by $z = 6$ with $w = 10$, and every value between them is available too. Only $20$ lies outside.

The circle of radius $r$ about $c$ is the set of $z$ with $|z - c| = r$, so the expression inside the bars has to be written as a difference before the center can be read.

$|z + 3 - 5i| = |z - (-3 + 5i)| = 6$

The center is $-3 + 5i$ and the radius is $6$.

Reduce the power first, then multiply. Since $51 = 4 \times 12 + 3$, the remainder is $3$ and $i^{51} = i^3 = -i$.

$-i(6 - 5i) = -6i + 5i^2 = -5 - 6i$

The $i^2$ is what puts a real part into a product that started with none: multiplying by $-i$ has moved the two parts and flipped a sign.

For a non-real pair $p \pm qi$ the product of the roots is $p^2 + q^2$, which is the squared modulus of either one, and the product is also $\tfrac{c}{a}$.

$r_1 r_2 = \frac{c}{a} = \frac{20}{5} = 4 = |z|^2 \quad\Longrightarrow\quad |z| = 2$

Every admissible $b$ moves the pair along the circle of radius $2$ about $0$ without changing that distance.