Complex Number Operations - Math 2

1. If $z_1 = 2 + 3i$ and $z_2 = 1 - 2i$, what is the coefficient of $i$ in the expansion of $(z_1 + z_2)^2$?
   Answer: $6$ First calculate $z_1 + z_2 = (2+3i) + (1-2i) = 3 + i$. Then $(z_1 + z_2)^2 = (3+i)^2 = 9 + 6i + i^2 = 9 + 6i - 1 = 8 + 6i$. The coefficient of $i$ is $6$.
2. Let $z = a + bi$ where $a$ and $b$ are real numbers. If $z^2 + z + 1 = 0$, what is the value of $|z|^2$?
   Answer: $1$ From $z^2 + z + 1 = 0$, we get $z^2 = -z - 1$. Multiplying both sides by $\bar{z}$ (the complex conjugate): $z^2\bar{z} = (-z-1)\bar{z} = -z\bar{z} - \bar{z} = -|z|^2 - \bar{z}$. Since $z^2\bar{z} = z \cdot z\bar{z} = z|z|^2$, we have $z|z|^2 = -|z|^2 - \bar{z}$. This gives $z|z|^2 + |z|^2 = -\bar{z}$, or $|z|^2(z + 1) = -\bar{z}$. Taking the conjugate of the original equation: $\overline{z^2 + z + 1} = \bar{z}^2 + \bar{z} + 1 = 0$. From $z^2 + z + 1 = 0$, we can use the quadratic formula: $z = \frac{-1 \pm \sqrt{1-4}}{2} = \frac{-1 \pm i\sqrt{3}}{2}$. For either root, $|z|^2 = \left(\frac{-1}{2}\right)^2 + \left(\pm\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1$.
3. Let $\alpha = \frac{1+i}{1-i}$ and $\beta = \frac{1-i}{1+i}$. What is the value of $\alpha^2 + \beta^2$?
   Answer: $-2$ This problem tests your ability to work with complex numbers and recognize important relationships between conjugate expressions. When you see complex fractions like these, start by simplifying each term. To simplify $\alpha = \frac{1+i}{1-i}$, multiply both numerator and denominator by the conjugate of the denominator: $\frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{(1+i)^2}{1-i^2} = \frac{1+2i+i^2}{1-(-1)} = \frac{1+2i-1}{2} = \frac{2i}{2} = i$. Similarly, for $\beta = \frac{1-i}{1+i}$, multiply by $\frac{1-i}{1-i}$: $\frac{(1-i)^2}{1+i^2} = \frac{1-2i+i^2}{2} = \frac{1-2i-1}{2} = \frac{-2i}{2} = -i$. Therefore, $\alpha^2 + \beta^2 = i^2 + (-i)^2 = -1 + (-1) = -2$. Choice A (-4) might result from incorrectly calculating $(2i)^2 + (-2i)^2$ without the division by 2. Choice B (2) could come from forgetting that $i^2 = -1$ and treating it as positive 1. Choice C (0) might occur if you mistakenly think $i^2$ and $(-i)^2$ cancel each other out rather than both equaling -1. When working with complex number ratios, always rationalize by multiplying by the conjugate of the denominator. Also remember that $i^2 = -1$, and $(-i)^2 = i^2 = -1$, not $-i^2$.
4. Let $u = 3 + 2i$ and $v = 1 - i$. If $uv + u\bar{v} = a + bi$ where $a$ and $b$ are real, what is $a + b$?
   Answer: $10$ First calculate $uv = (3+2i)(1-i) = 3 - 3i + 2i - 2i^2 = 3 - i + 2 = 5 - i$. Next, $\bar{v} = \overline{1-i} = 1+i$, so $u\bar{v} = (3+2i)(1+i) = 3 + 3i + 2i + 2i^2 = 3 + 5i - 2 = 1 + 5i$. Therefore, $uv + u\bar{v} = (5-i) + (1+5i) = 6 + 4i$. So $a = 6$ and $b = 4$, giving $a + b = 10$.
5. Let $w = 2 + 3i$ and $v = 1 - 2i$. If $p = w \cdot \overline{v} + \overline{w} \cdot v$, what type of number is $p$?
   Answer: A real number We have $w = 2 + 3i$, so $\overline{w} = 2 - 3i$. Also $v = 1 - 2i$, so $\overline{v} = 1 + 2i$. Now compute $w \cdot \overline{v} = (2+3i)(1+2i) = 2 + 4i + 3i + 6i^2 = 2 + 7i - 6 = -4 + 7i$. And $\overline{w} \cdot v = (2-3i)(1-2i) = 2 - 4i - 3i + 6i^2 = 2 - 7i - 6 = -4 - 7i$. Therefore $p = (-4+7i) + (-4-7i) = -8$. Since $p = -8$ is a real number, the answer is C. Notice that $p = w\overline{v} + \overline{w}v = w\overline{v} + \overline{w\overline{v}}$, which is always real because it has the form $z + \overline{z}$ for $z = w\overline{v}$. Choice A and B would appeal to students who make computational errors. Choice D would appeal to students who don't recognize the general pattern.
6. If $w_1 = 4 - 3i$ and $w_2 = -1 + 2i$, which expression represents the conjugate of $w_1w_2$?
   Answer: $\overline{w_1} \cdot \overline{w_2}$ This tests the property that the conjugate of a product equals the product of the conjugates: $\overline{w_1w_2} = \overline{w_1} \cdot \overline{w_2}$. To verify: $w_1w_2 = (4-3i)(-1+2i) = -4 + 8i + 3i - 6i^2 = -4 + 11i + 6 = 2 + 11i$, so $\overline{w_1w_2} = 2 - 11i$. Also, $\overline{w_1} = 4 + 3i$ and $\overline{w_2} = -1 - 2i$, giving $\overline{w_1} \cdot \overline{w_2} = (4+3i)(-1-2i) = -4 - 8i - 3i - 6i^2 = -4 - 11i + 6 = 2 - 11i$. These match, confirming the property.
7. For complex numbers $s = 1 + 2i$ and $t = 3 - i$, which of the following equals $|st|^2$?
   Answer: $|s|^2 \cdot |t|^2$ When you encounter problems involving the modulus (absolute value) of products of complex numbers, remember that there's a fundamental property that makes these calculations much simpler than computing the product first. The key insight is that $|st| = |s| \cdot |t|$ for any complex numbers $s$ and $t$. Therefore, $|st|^2 = (|s| \cdot |t|)^2 = |s|^2 \cdot |t|^2$. Let's verify this with our given values. First, find the moduli: $|s| = |1 + 2i| = \sqrt{1^2 + 2^2} = \sqrt{5}$ and $|t| = |3 - i| = \sqrt{3^2 + (-1)^2} = \sqrt{10}$. So $|s|^2 \cdot |t|^2 = 5 \cdot 10 = 50$. Now let's see why the other options fail. Choice A gives $|s|^2 + |t|^2 = 5 + 10 = 15$, which confuses addition with multiplication. Choice B yields $(|s| - |t|)^2 = (\sqrt{5} - \sqrt{10})^2$, which is negative since $\sqrt{5} < \sqrt{10}$, and squares of real numbers can't be negative. Choice C gives $(|s| + |t|)^2 = (\sqrt{5} + \sqrt{10})^2 = 5 + 10 + 2\sqrt{50} = 15 + 10\sqrt{2}$, which incorrectly treats the modulus of a product like the modulus of a sum. The correct answer is D: $|s|^2 \cdot |t|^2$. Study tip: Memorize that $|zw| = |z| \cdot |w|$ for complex numbers. This property often eliminates the need to multiply complex numbers explicitly, saving time and reducing errors.
8. If $z = \frac{2+i}{1-2i}$, what is the value of $z + \bar{z}$?
   Answer: $0$ To find $z$, multiply numerator and denominator by the conjugate of the denominator: $z = \frac{2+i}{1-2i} \cdot \frac{1+2i}{1+2i} = \frac{(2+i)(1+2i)}{(1-2i)(1+2i)} = \frac{2+4i+i+2i^2}{1+4} = \frac{2+5i-2}{5} = \frac{5i}{5} = i$. Therefore, $\bar{z} = \overline{i} = -i$, and $z + \bar{z} = i + (-i) = 0$.
9. Let $a$ and $b$ be real numbers such that $(a + bi)^2 = -5 + 12i$. What is the value of $a^2 + b^2$? Give your answer in simplest form.
   Answer: $13$ Expanding $(a+bi)^2 = a^2 + 2abi + (bi)^2 = a^2 - b^2 + 2abi = -5 + 12i$. Equating real and imaginary parts: $a^2 - b^2 = -5$ and $2ab = 12$, so $ab = 6$. We need to find $a^2 + b^2$. From $ab = 6$, we get $b = \frac{6}{a}$. Substituting into the first equation: $a^2 - \frac{36}{a^2} = -5$. Multiplying by $a^2$: $a^4 + 5a^2 - 36 = 0$. Let $u = a^2$: $u^2 + 5u - 36 = 0$. Factoring: $(u+9)(u-4) = 0$. Since $u = a^2 \geq 0$, we have $u = 4$, so $a^2 = 4$. Then $b^2 = \frac{36}{a^2} = \frac{36}{4} = 9$. Therefore $a^2 + b^2 = 4 + 9 = 13$. Choice A gives the modulus of the original complex number, choice C is just the real part's absolute value, and choice D is $\sqrt{169} = 13$ but in the wrong form.
10. If $z = a + bi$ where $a$ and $b$ are real, and $z \cdot \overline{z} + z + \overline{z} = 7$, what is the value of $a^2 + b^2 + 2a$?
    Answer: $7$ We have $z = a + bi$ and $\overline{z} = a - bi$. First, $z \cdot \overline{z} = (a+bi)(a-bi) = a^2 + b^2$. Also, $z + \overline{z} = (a+bi) + (a-bi) = 2a$. So the equation becomes $a^2 + b^2 + 2a = 7$. Therefore, $a^2 + b^2 + 2a = 7$. The answer is directly given by the constraint equation. Choice A might come from students who forget the factor of 2 in $z + \overline{z}$. Choice C might result from adding an extra 2 somewhere. Choice D might come from computational errors in expanding the products.

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