The Imaginary Unit and Complex Numbers: Additional Practice (Set 1)

Only the remainder of the exponent on division by $4$ matters, and since $100$ is a multiple of $4$, the last two digits decide it: $01$ leaves remainder $1$.

$i^{2101} = \left(i^4\right)^{525} \cdot i^1 = 1 \cdot i = i$

Remainder $1$ lands on $i$, the first stop of the cycle.

Apply the definition $\sqrt{-k} = i\sqrt{k}$ with $k = 121$.

$\sqrt{-121} = i\sqrt{121} = 11i$

The radical names one number, $11i$; the pair $\pm 11i$ is the solution set of the equation $x^2 = -121$, which is a different object.

Convert to $i$-form, then factor out the largest perfect square, $48 = 16 \cdot 3$.

$\sqrt{-48} = i\sqrt{48} = i \cdot 4\sqrt{3} = 4i\sqrt{3}$

Check: $\left(4i\sqrt{3}\right)^2 = 16 \cdot (-1) \cdot 3 = -48$.

Reduce each exponent by its remainder on division by $4$: $7 = 4 + 3$ gives $i^7 = i^3 = -i$, and $9 = 8 + 1$ gives $i^9 = i$.

$i^7 + i^9 = -i + i = 0$

The two powers are opposites, so the sum vanishes.

Write the number in standard form: $3 - i = 3 + (-1)i$.

$\text{imaginary part} = -1$

The imaginary part is the real coefficient of $i$, sign included, so it is $-1$, not $-i$ and not $1$.

Match the parts: the real parts give $x + y = 10$ and the imaginary parts give $x - y = 2$.

$2x = (x + y) + (x - y) = 12 \quad\Rightarrow\quad x = 6$

Adding the two equations eliminates $y$; then $y = 4$, and the check $(6 + 4) + (6 - 4)i = 10 + 2i$ confirms the pair.

Evaluate each term from the cycle: $i$, then $-1$, then $-i$.

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

The $i$ and $-i$ cancel, leaving $-1$. The sum only reaches zero once the fourth power, $+1$, joins in.

Pure imaginary means the real part is zero while the imaginary part is not.

$k - 2 = 0 \quad\Rightarrow\quad k = 2$

The imaginary part is already $7 \ne 0$, so $k = 2$ makes the number $0 + 7i = 7i$, which is pure imaginary.

A complex number is real exactly when its imaginary part is zero.

$m + 5 = 0 \quad\Rightarrow\quad m = -5$

With $m = -5$ the number becomes $4 + 0i = 4$, a plain real number.

Reduce each exponent by its remainder on division by $4$: the exponents $4$, $8$, and $12$ all leave remainder $0$, giving $1$.

$i^6 = i^4 \cdot i^2 = 1 \cdot (-1) = -1$

Only $i^6$, with remainder $2$, lands on $-1$. An even exponent is not enough; the remainder must be $2$.

Convert the numerator to $i$-form first: $\sqrt{-27} = i\sqrt{27} = 3i\sqrt{3}$.

$\frac{3i\sqrt{3}}{\sqrt{3}} = 3i$

The factor $\sqrt{3}$ cancels top and bottom, leaving the pure imaginary number $3i$.

Write $n = 4q$ for a positive integer $q$ and use $i^4 = 1$.

$i^n = \left(i^4\right)^q = 1^q = 1$

A multiple of $4$ means a whole number of complete cycles, and each complete cycle multiplies to $1$.