The Imaginary Unit and Complex Numbers: Practice

Only the remainder of the exponent on division by $4$ matters, and since $100$ is a multiple of $4$, only the last two digits decide it: $27 = 4 \cdot 6 + 3$.

$i^{2027} = \left(i^4\right)^{506} \cdot i^3 = 1 \cdot i^3 = -i$

Remainder $3$ lands on $-i$ in the cycle $i, -1, -i, 1$.

Each block of four consecutive powers is one full cycle, and one cycle sums to zero.

$i + i^2 + i^3 + i^4 = i - 1 - i + 1 = 0$

The $100$ terms split into exactly $25$ such blocks, so the whole sum is $25 \cdot 0 = 0$.

The first $100$ terms form $25$ complete cycles and contribute $0$, leaving only the last two terms.

$i^{101} + i^{102} = i + (-1) = -1 + i$

Here $101 = 4 \cdot 25 + 1$ gives $i^{101} = i$, and $102 = 4 \cdot 25 + 2$ gives $i^{102} = -1$.

Isolate the square first.

$x^2 = -\frac{49}{4} \quad\Rightarrow\quad x = \pm i\sqrt{\tfrac{49}{4}} = \pm \tfrac{7}{2} i$

Check: $\left(\tfrac{7}{2}i\right)^2 = \tfrac{49}{4}\,i^2 = -\tfrac{49}{4}$, so $4x^2 = -49$ and $4x^2 + 49 = 0$.

Raise each factor to the fourth power separately.

$(2i)^4 = 2^4 \cdot i^4 = 16 \cdot 1 = 16$

Since $i^4 = 1$, the answer is real and positive; the fourth power wipes out every trace of $i$.

Equal complex numbers match part by part, so the one equation splits into two real ones.

$2x + 1 = 7 \quad\Rightarrow\quad x = 3 \qquad\qquad y - 3 = -5 \quad\Rightarrow\quad y = -2$

Check: $(2 \cdot 3 + 1) + (-2 - 3)i = 7 - 5i$, as required.

Zero is the complex number $0 + 0i$, so both parts must vanish at once: $x^2 - 9 = 0$ and $x + 3 = 0$.

$x^2 = 9 \ \text{ gives } \ x = \pm 3, \qquad x + 3 = 0 \ \text{ gives } \ x = -3$

Only $x = -3$ satisfies both conditions; $x = 3$ kills the real part but leaves the imaginary part $6i$ standing.

The exponent $4n + 2$ leaves remainder $2$ on division by $4$, whatever $n$ is.

$i^{4n+2} = \left(i^4\right)^n \cdot i^2 = 1^n \cdot (-1) = -1$

Every such power lands on the second stop of the cycle.

Add the exponents, then reduce by the cycle.

$i^{45} \cdot i^{23} = i^{68} = \left(i^4\right)^{17} = 1$

Since $68 = 4 \cdot 17$ exactly, the remainder is $0$ and the product is $1$.

Multiply the exponents, then reduce: $15 = 4 \cdot 3 + 3$, remainder $3$.

$\left(i^3\right)^5 = i^{15} = \left(i^4\right)^3 \cdot i^3 = i^3 = -i$

Alternatively, $i^3 = -i$, and $(-i)^5 = (-1)^5 i^5 = -i^5 = -i$, the same answer.

Isolate the square, then simplify the radical using $12 = 4 \cdot 3$.

$x^2 = -12 \quad\Rightarrow\quad x = \pm i\sqrt{12} = \pm 2i\sqrt{3}$

Check: $\left(2i\sqrt{3}\right)^2 = 4 \cdot (-1) \cdot 3 = -12$.

Square the product: $(bi)^2 = b^2 i^2 = -b^2$, so the condition becomes $-b^2 = -81$.

$b^2 = 81 \quad\Rightarrow\quad b = \pm 9$

Both signs work, since squaring erases the sign of $b$. The option $\pm 9i$ is ruled out because the problem requires $b$ to be real.