The Imaginary Unit and Complex Numbers: Additional Practice (Set 2)

Rewrite as $x^2 = -k$. A real solution exists exactly when the right side is nonnegative, since real squares are exactly the nonnegative numbers.

$-k \ge 0 \quad\Longleftrightarrow\quad k \le 0$

For $k = 0$ the solution is $x = 0$; for $k < 0$ the solutions are $x = \pm\sqrt{-k}$, real because $-k > 0$; for $k > 0$ the solutions $\pm i\sqrt{k}$ are not real.

Isolate the square: $x^2 = -100$. Test the candidates suggested by $\sqrt{-100} = 10i$.

$(10i)^2 = 100\,i^2 = -100 \qquad (-10i)^2 = 100\,i^2 = -100$

Both check, so the solutions are the mirror pair $x = \pm 10i$.

Convert each radical to $i$-form and simplify: $\sqrt{-50} = i\sqrt{50} = 5i\sqrt{2}$ and $\sqrt{-8} = i\sqrt{8} = 2i\sqrt{2}$.

$5i\sqrt{2} + 2i\sqrt{2} = 7i\sqrt{2}$

The terms combine because both are multiples of the same quantity $i\sqrt{2}$. Adding the radicands to get $\sqrt{58}\,i$ is not a legal move for square roots.

Convert each factor to $i$-form first; the radical product rule does not cover two negative radicands.

$\sqrt{-5} \cdot \sqrt{-20} = \left(i\sqrt{5}\right)\left(i\sqrt{20}\right) = i^2 \sqrt{100} = -10$

The real radicands $5$ and $20$ multiply legally to $100$, and the $i^2$ supplies the minus sign that the naive rule misses.

Convert to $i$-form first; the product rule for radicals does not apply to two negative radicands.

$\sqrt{-6} \cdot \sqrt{-6} = \left(i\sqrt{6}\right)^2 = i^2 \cdot 6 = -6$

This is also forced by the meaning of a square root: $\sqrt{-6}$ is defined as a number whose square is $-6$, so squaring it must return $-6$, not $6$.

Square each factor separately.

$(5i)^2 = 25\,i^2 = 25 \cdot (-1) = -25$

Squaring a pure imaginary number always produces a negative real number; forgetting the $i^2 = -1$ is what produces $25$.

Cube each factor separately, using $i^3 = -i$ from the cycle.

$(3i)^3 = 3^3 \cdot i^3 = 27 \cdot (-i) = -27i$

The answer is pure imaginary: an odd power of a pure imaginary number always is.

Raise each factor to the fifth power, and reduce $i^5$ by the cycle: $5 = 4 + 1$, so $i^5 = i$.

$(2i)^5 = 2^5 \cdot i^5 = 32 \cdot i = 32i$

The answer is pure imaginary, as every odd power of a pure imaginary number is.

A real number $a$ is the complex number $a + 0i$, so the first statement is true, and by definition the imaginary part is the real coefficient $b$, so the third is true. The equation $x^2 = -1$ has exactly the pair $\pm i$, so the fourth is true.

$i = 0 + 1i \ \text{ is complex but not real}$

So the false statement is that every complex number is real: the inclusion runs in one direction only.

The solutions of $x^2 = -k$ are $x = \pm i\sqrt{k}$, and $\sqrt{196} = 14$.

$x = \pm i\sqrt{196} = \pm 14i$

Check: $(\pm 14i)^2 = 196\,i^2 = -196$, as required.

Raise each to the fourth power: $1^4 = 1$ and $(-1)^4 = 1$ are immediate, and the cycle gives $i^4 = 1$.

$(-i)^4 = (-1)^4 \cdot i^4 = 1 \cdot 1 = 1$

So all four numbers are solutions of $x^4 = 1$. Enlarging the number system gave this equation four roots where the reals could only supply two.

Square each factor separately.

$\left(i\sqrt{7}\right)^2 = i^2 \cdot \left(\sqrt{7}\right)^2 = (-1) \cdot 7 = -7$

This is exactly the check that makes the definition $\sqrt{-7} = i\sqrt{7}$ legitimate: the candidate really does square to $-7$.