Question

The square roots of the complex number $(-5 - 12i)$ are

• A. $\pm (2 - 3i)$
• B. $\pm (3 + 2i)$
• C. $\pm (2 + 3i)$
• D. $\pm (3 - 2i)$

Hint:

Bypass long algebraic expansions entirely during multiple-choice tests by squaring the options! For option (A): $(\pm(2-3i))^2 = 4 + 9i^2 - 12i = 4 - 9 - 12i = -5 - 12i$. Since this squares back to our original number instantly, it confirms the answer in seconds.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem requires finding the algebraic square roots of the given complex number $z = -5 - 12i$.

Step 2: Key Formula or Approach:
Let the square root of the complex number be $a + ib$. Therefore: $$\sqrt{-5 - 12i} = a + ib \implies -5 - 12i = (a + ib)^2$$ Expanding the right-hand side gives: $$-5 - 12i = (a^2 - b^2) + i(2ab)$$ By equating the real parts and imaginary parts separately, we set up a system of quadratic equations to determine real values for $a$ and $b$.

Step 3: Detailed Explanation:
Equating matching components from our expansion: align a^2 - b^2 &= -5
2ab &= -12 ab = -6 align From equation (2), we isolate $b$: $b = -\frac{6}{a}$. Substitute this expression into equation (1): $$a^2 - \left(-\frac{6}{a}\right)^2 = -5 \implies a^2 - \frac{36}{a^2} = -5$$ Multiply the entire equation by $a^2$ to form a quadratic equation in terms of $a^2$: $$a^4 - 36 = -5a^2 \implies a^4 + 5a^2 - 36 = 0$$ Factor the bi-quadratic equation: $$(a^2 + 9)(a^2 - 4) = 0$$ Since $a$ must be a real number, $a^2$ cannot be negative ($a^2 = -9$ is rejected). Thus: $$a^2 = 4 \implies a = \pm 2$$ Now, substitute these values back to solve for $b$ using $b = -\frac{6}{a}$:

• If $a = 2$, then $b = -\frac{6}{2} = -3 \implies z^{1/2} = 2 - 3i$

• If $a = -2$, then $b = -\frac{6}{-2} = 3 \implies z^{1/2} = -2 + 3i = -(2 - 3i)$
Combining these solutions gives $\pm(2 - 3i)$.

Step 4: Final Answer:
The square roots are $\pm (2 - 3i)$, which corresponds to option (A).