Solving the Complex Roots Equation
We are given the set $S = \{z \in \mathbb{C} : z^2 + 4z + 16 = 0\}$. First, we need to find the roots $z$ of the quadratic equation $z^2 + 4z + 16 = 0$. We use the quadratic formula: $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For this equation, $a=1$, $b=4$, and $c=16$. The discriminant is $\Delta = b^2 - 4ac = 4^2 - 4(1)(16) = 16 - 64 = -48$. So, the roots are: $z = \frac{-4 \pm \sqrt{-48}}{2(1)} = \frac{-4 \pm i\sqrt{48}}{2} = \frac{-4 \pm i(4\sqrt{3})}{2} = -2 \pm 2\sqrt{3}i$ The roots are $z_1 = -2 + 2\sqrt{3}i$ and $z_2 = -2 - 2\sqrt{3}i$. Thus, $S = \{-2 + 2\sqrt{3}i, -2 - 2\sqrt{3}i\}$.
Calculating $|z + \sqrt{3}i|^2$ for Each Root
Next, we calculate the value of $|z + \sqrt{3}i|^2$ for each root in the set $S$. Let $w = \sqrt{3}i$. We need to find $|z + w|^2$. For $z_1 = -2 + 2\sqrt{3}i$: $z_1 + w = (-2 + 2\sqrt{3}i) + \sqrt{3}i = -2 + (2\sqrt{3} + \sqrt{3})i = -2 + 3\sqrt{3}i$. The modulus squared $|a+bi|^2$ is $a^2 + b^2$. $|z_1 + w|^2 = |-2 + 3\sqrt{3}i|^2 = (-2)^2 + (3\sqrt{3})^2 = 4 + (9 \times 3) = 4 + 27 = 31$. For $z_2 = -2 - 2\sqrt{3}i$: $z_2 + w = (-2 - 2\sqrt{3}i) + \sqrt{3}i = -2 + (-2\sqrt{3} + \sqrt{3})i = -2 - \sqrt{3}i$. $|z_2 + w|^2 = |-2 - \sqrt{3}i|^2 = (-2)^2 + (-\sqrt{3})^2 = 4 + 3 = 7$.
Summing the Modulus Squared Values
Finally, we sum the calculated values: $\sum_{z \in S} |z + \sqrt{3}i|^2 = |z_1 + \sqrt{3}i|^2 + |z_2 + \sqrt{3}i|^2$ Sum $= 31 + 7 = 38$. The required sum is 38.
Similar Questions
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Important Questions from Algebra
Let A = {-3, -2, -1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if $0\le x^2+2y\le 4$. Let $l$ be the number of elements in R and $m$ be the minimum number of elements required to be added in R to make it a reflexive relation. Then $l + m$ is equal to
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