Complex Numbers and De Moivre’s Theorem — Flashcards

Answer

Step 1: $\sqrt{-48} = \sqrt{48 \cdot (-1)} = \sqrt{48} \cdot \sqrt{-1}$ Step 2: $\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3}$ Step 3: $\sqrt{-1} = i$ Therefore, $\sqrt{-48} = 4\sqrt{3}i$…

Answer

Step 1: $\sqrt{-5} = \sqrt{5}i$, $\sqrt{-20} = \sqrt{20}i = 2\sqrt{5}i$ Step 2: Multiply: $(\sqrt{5}i)(2\sqrt{5}i) = 2 \cdot 5 \cdot i^2 = 10 \cdot (-1) = -10$ Note: $\sqrt{a} \cdot \sqrt{b} = \sqrt…

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Step 1: Divide 27 by 4 → Quotient = 6, Remainder = 3 Step 2: $i^{27} = i^{4\cdot6 + 3} = (i^4)^6 \cdot i^3 = (1)^6 \cdot i^3 = i^3$ Step 3: $i^3 = i^2 \cdot i = (-1) \cdot i = -i$ Answer: $-i$…

Answer

Step 1: $i^{10} = (i^2)^5 = (-1)^5 = -1$ Step 2: $i^{20} = (i^2)^{10} = (-1)^{10} = 1$ Step 3: $i^{30} = (i^2)^{15} = (-1)^{15} = -1$ Step 4: Add: $1 + (-1) + 1 + (-1) = 0$ Answer: $0$…

Answer

Step 1: $i^3 = -i$, $i^{16} = (i^4)^4 = 1$, $i^{11} = i^{8+3} = (i^4)^2 \cdot i^3 = 1 \cdot (-i) = -i$ Step 2: $8(-i) + 6(1) - 12(-i) = -8i + 6 + 12i$ Step 3: Combine: $6 + 4i$ Answer: $6 + 4i$…

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The conjugate is obtained by changing the sign of the imaginary part. So, conjugate of $3 - 4i$ is $3 + 4i$.

Answer

Step 1: Expand $(2 + i)^2 = 4 + 4i + i^2 = 4 + 4i - 1 = 3 + 4i$ Step 2: Conjugate of $3 + 4i$ is $3 - 4i$ Answer: $3 - 4i$…

Answer

Modulus $|z| = \sqrt{a^2 + b^2}$ where $z = a + bi$ Here, $a = -4$, $b = 3$ $|z| = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ Answer: $5$…