To find the square root of the complex number $-48 - 14i$, we will express the result in the standard form $a + bi$, where $a$ and $b$ are real numbers.

Let's go through the steps:

1. Set up the equation: $\sqrt{-48 - 14i} = a + bi$ where $a$ and $b$ are real numbers.

2. Square both sides: $(a + bi)^2 = -48 - 14i$ Expanding the left side using the identity $(a + bi)^2 = a^2 - b^2 + 2abi$, we get: $a^2 - b^2 + 2abi = -48 - 14i$

3. Equate real and imaginary parts: From the above, equate the real and imaginary parts separately:

   • Real part: $a^2 - b^2 = -48$
   • Imaginary part: $2ab = -14$

4. Solve the system of equations:

   From the imaginary part equation $2ab = -14$, we can solve for $ab$: $ab = -7$

   Now, substitute $b = -\frac{7}{a}$ into the real part equation: $a^2 - \left( -\frac{7}{a} \right)^2 = -48$ Simplify: $a^2 - \frac{49}{a^2} = -48$

   Multiply both sides by $a^2$ to eliminate the fraction: $a^4 + 48a^2 - 49 = 0$

   Let $x = a^2$, so we have the quadratic equation: $x^2 + 48x - 49 = 0$

5. Solve the quadratic equation: Using the quadratic formula: $x = \frac{-48 \pm \sqrt{48^2 - 4(1)(-49)}}{2(1)}$ Simplify: $x = \frac{-48 \pm \sqrt{2304 + 196}}{2}$ $x = \frac{-48 \pm \sqrt{2500}}{2}$ $x = \frac{-48 \pm 50}{2}$ So, $x = \frac{-48 + 50}{2} = 1 \quad \text{or} \quad x = \frac{-48 - 50}{2} = -49$ Since $x = a^2 \geq 0$, we discard $x = -49$ and take $x = 1$. Therefore, $a^2 = 1$, so $a = 1$ or $a = -1$.

6. Find $b$: Using $ab = -7$ and $a = 1$ or $a = -1$, we can find $b$:

   • If $a = 1$, then $b = -7$.
   • If $a = -1$, then $b = 7$.

7. Final solution: Therefore, the square roots of $-48 - 14i$ are: $\boxed{1 - 7i \quad \text{or} \quad -1 + 7i}$

Would you like more details on any step, or further clarification?

---

Here are 5 related questions:

1. How do you determine the polar form of a complex number?
2. What is the geometric interpretation of the square root of a complex number?
3. How can we apply De Moivre’s Theorem to find roots of complex numbers?
4. How does the discriminant in quadratic equations affect the nature of roots?
5. What are the applications of complex numbers in physics and engineering?

Tip: When solving for square roots of complex numbers, expressing them in polar form can sometimes simplify the process!