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Question

Let $z_{1}$, $z_{2} \in \mathbb{C}$ be the distinct solutions of the equation $z^{2}+4z-(1+12i)=0$. Then $|z_{1}|^{2}+|z_{2}|^{2}$ is equal to :

The correct answer is
34

Calculating Sum of Squared Magnitudes of Complex Roots

The problem requires finding the value of $|z_1|^2 + |z_2|^2$, where $z_1$ and $z_2$ are the distinct complex solutions to the quadratic equation $z^2 + 4z - (1 + 12i) = 0$. The most direct method is to find the roots.

Step 1: Identify Equation Coefficients

For the quadratic equation $az^2 + bz + c = 0$, we have:

  • $a = 1$
  • $b = 4$
  • $c = -(1 + 12i)$

Step 2: Calculate the Discriminant

The discriminant $\Delta$ is calculated using $\Delta = b^2 - 4ac$.

$ \Delta = 4^2 - 4(1)(-(1 + 12i)) $

$ \Delta = 16 + 4(1 + 12i) $

$ \Delta = 16 + 4 + 48i $

$ \Delta = 20 + 48i $

Step 3: Find the Square Root of the Discriminant

We need to find $\sqrt{20 + 48i}$. Let $\sqrt{20 + 48i} = x + yi$, where $x, y \in \mathbb{R}$.

Squaring gives $(x + yi)^2 = x^2 - y^2 + 2xyi = 20 + 48i$. Equating real and imaginary parts yields:

  • $x^2 - y^2 = 20$ (Eq. 1)
  • $2xy = 48 \implies xy = 24$ (Eq. 2)

We also use $|x+yi|^2 = |20+48i|$, which gives $x^2 + y^2 = \sqrt{20^2 + 48^2} = \sqrt{400 + 2304} = \sqrt{2704} = 52$ (Eq. 3).

Adding Eq. 1 and Eq. 3: $2x^2 = 72 \implies x^2 = 36 \implies x = \pm 6$.

Subtracting Eq. 1 from Eq. 3: $2y^2 = 32 \implies y^2 = 16 \implies y = \pm 4$.

Since $xy = 24$ (positive), $x$ and $y$ must have the same sign. Therefore, $\sqrt{20 + 48i} = \pm (6 + 4i)$.

Step 4: Find the Complex Roots ($z_1, z_2$)

Using the quadratic formula $z = \frac{-b \pm \sqrt{\Delta}}{2a}$:

$ z = \frac{-4 \pm (6 + 4i)}{2} $

The distinct roots are:

  • $z_1 = \frac{-4 + (6 + 4i)}{2} = \frac{2 + 4i}{2} = 1 + 2i$
  • $z_2 = \frac{-4 - (6 + 4i)}{2} = \frac{-10 - 4i}{2} = -5 - 2i$

Step 5: Calculate the Squared Magnitudes

The squared magnitude $|z|^2$ for $z = x + yi$ is $x^2 + y^2$.

$ |z_1|^2 = |1 + 2i|^2 = 1^2 + 2^2 = 1 + 4 = 5 $

$ |z_2|^2 = |-5 - 2i|^2 = (-5)^2 + (-2)^2 = 25 + 4 = 29 $

Step 6: Compute the Sum of Squared Magnitudes

The required sum is $|z_1|^2 + |z_2|^2$.

$ |z_1|^2 + |z_2|^2 = 5 + 29 = 34 $

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Similar Questions

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Important Questions from Algebra

  1. Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
  2. Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
  3. Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
  4. If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
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