The Imaginary Unit and Complex Numbers: Practice

Divide the exponent by $4$ with remainder: $10 = 4 \cdot 2 + 2$.

$i^{10} = \left(i^4\right)^2 \cdot i^2 = 1 \cdot i^2 = -1$

Only the remainder matters, and remainder $2$ lands on $-1$ in the cycle $i, -1, -i, 1$.

Divide the exponent by $4$ with remainder: $27 = 4 \cdot 6 + 3$.

$i^{27} = \left(i^4\right)^6 \cdot i^3 = 1 \cdot i^3 = -i$

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

Here $100 = 4 \cdot 25$ exactly, so the remainder is $0$.

$i^{100} = \left(i^4\right)^{25} = 1^{25} = 1$

A remainder of $0$ means the exponent is a whole number of complete cycles, and each complete cycle multiplies to $1$.

Isolate the square: $x^2 = -64$. No real number works, so test the candidates suggested by $\sqrt{-64} = 8i$.

$(8i)^2 = 64\,i^2 = -64 \qquad (-8i)^2 = 64\,i^2 = -64$

Both check, so the complete answer is the pair $x = \pm 8i$.

Divide by $2$ first, then isolate the square.

$2x^2 = -50 \quad\Rightarrow\quad x^2 = -25$

The solutions are $x = \pm i\sqrt{25} = \pm 5i$, and each checks: $(\pm 5i)^2 = 25\,i^2 = -25$.

Convert both factors to $i$-form before multiplying.

$\sqrt{-2} \cdot \sqrt{-8} = \left(i\sqrt{2}\right)\left(i\sqrt{8}\right) = i^2 \sqrt{16} = -4$

The real radicals multiply legally because $2$ and $8$ are nonnegative; the $i^2$ then contributes the minus sign.

Equal complex numbers have equal real parts and equal imaginary parts. The imaginary parts already agree, so match the real parts.

$x + 2 = 7 \quad\Rightarrow\quad x = 5$

One complex equation carries two real equations, and here only the real-part equation has any work in it.

Add the exponents first, since the exponent laws still hold for $i$.

$i^3 \cdot i^5 = i^8 = \left(i^4\right)^2 = 1$

Alternatively, $i^3 = -i$ and $i^5 = i$, and $(-i)(i) = -i^2 = 1$, the same answer.

Evaluate each power from the cycle: $i^2 = -1$ and $i^4 = 1$.

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

The two powers are exact opposites, so the sum collapses to zero.

Rearrange the factors, then collapse $i^2$.

$5i \cdot 4i = 20\,i^2 = 20 \cdot (-1) = -20$

Forgetting the $i^2 = -1$ step is what produces $20$ or $20i$.

The solutions of $x^2 = -k$ are $x = \pm i\sqrt{k}$, so simplify $\sqrt{20}$ with $20 = 4 \cdot 5$.

$x = \pm i\sqrt{20} = \pm 2i\sqrt{5}$

Check: $\left(2i\sqrt{5}\right)^2 = 4 \cdot (-1) \cdot 5 = -20$.

Match the parts separately. The real parts give $a - 1 = 6$, so $a = 7$; the imaginary parts give $b = 3$.

$a + b = 7 + 3 = 10$

Equality of complex numbers always splits into exactly these two real equations.