Key Concepts and Formulas

• Modulus of a Complex Number: For $z = x + iy$, $|z| = \sqrt{x^2 + y^2}$ and $|z| \ge 0$.
• Logarithm Properties: $b \log_a x = \log_a x^b$, $e^{\log_e A} = A$, $\log_a b = k \iff a^k = b$. If $a>1$ and $a^x \ge a^y$, then $x \ge y$.
• Solving Inequalities: Rational inequalities are solved by moving all terms to one side, finding a common denominator, and analyzing the sign in different intervals based on the roots of the numerator and denominator.

Step-by-Step Solution

Step 1: Substitute |z| with t

To simplify the expression, let $t = |z|$. Explanation: This substitution converts the problem from complex numbers to real numbers, making it easier to manipulate algebraically.

$t = |z|$

Step 2: Establish the Domain of t

Since the modulus of any complex number is non-negative, $t \ge 0$. Explanation: This is a fundamental property of the modulus and is crucial for determining the valid range of solutions.

$t \ge 0$

Step 3: Simplify the Absolute Value in the Denominator

The expression $||z| + 1|$ becomes $|t + 1|$. Since $t \ge 0$, $t + 1 \ge 1 > 0$, so $|t + 1| = t + 1$. Explanation: Because we know $t$ is non-negative, $t+1$ is always positive, so the absolute value signs can be removed.

$|t+1| = t+1$

Step 4: Evaluate the Right-Hand Side (RHS)

The RHS is $\log_{\sqrt{2}} |5\sqrt{7} + 9i|$. First, we find the modulus of $5\sqrt{7} + 9i$: $|5\sqrt{7} + 9i| = \sqrt{(5\sqrt{7})^2 + 9^2} = \sqrt{25 \cdot 7 + 81} = \sqrt{175 + 81} = \sqrt{256} = 16$ Now, we evaluate $\log_{\sqrt{2}} 16$. Since $16 = 2^4 = (\sqrt{2})^8$, we have: $\log_{\sqrt{2}} 16 = \log_{\sqrt{2}} (\sqrt{2})^8 = 8$ Explanation: This step calculates the modulus of the complex number and then evaluates the logarithm. We rewrite 16 as a power of $\sqrt{2}$ to easily evaluate the logarithm.

$|5\sqrt{7} + 9i| = 16$ $\log_{\sqrt{2}} 16 = 8$

Step 5: Simplify the Left-Hand Side (LHS)

The LHS is $\exp \left( \frac{(|z| + 3)(|z| - 1)}{||z| + 1|} \log_e 2 \right)$. Substituting $t = |z|$ and simplifying the absolute value, we get: $\exp \left( \frac{(t + 3)(t - 1)}{t + 1} \log_e 2 \right) = e^{\frac{(t + 3)(t - 1)}{t + 1} \ln 2}$ Using the logarithm property $b \log_a x = \log_a x^b$, we get: $e^{\ln 2^{\frac{(t + 3)(t - 1)}{t + 1}}}$ Using the inverse property $e^{\ln A} = A$, we get: $2^{\frac{(t + 3)(t - 1)}{t + 1}}$ Explanation: This step uses the properties of logarithms and exponentials to simplify the LHS. The key is using the power rule of logarithms and the inverse relationship between $e$ and $\ln$ to eliminate the exponential and logarithmic terms.

$e^{\frac{(t + 3)(t - 1)}{t + 1} \ln 2} = 2^{\frac{(t + 3)(t - 1)}{t + 1}}$

Step 6: Reformulate and Solve the Inequality

The inequality is now: $2^{\frac{(t + 3)(t - 1)}{t + 1}} \ge 8$ Since $8 = 2^3$, we have: $2^{\frac{(t + 3)(t - 1)}{t + 1}} \ge 2^3$ Since the base is 2 (which is greater than 1), we can compare the exponents directly: $\frac{(t + 3)(t - 1)}{t + 1} \ge 3$ $\frac{(t + 3)(t - 1)}{t + 1} - 3 \ge 0$ $\frac{(t + 3)(t - 1) - 3(t + 1)}{t + 1} \ge 0$ $\frac{t^2 + 2t - 3 - 3t - 3}{t + 1} \ge 0$ $\frac{t^2 - t - 6}{t + 1} \ge 0$ Explanation: This step rewrites the inequality with the same base on both sides, allowing us to compare the exponents. Then, we manipulate the algebraic expression to get a single fraction on one side.

$2^{\frac{(t + 3)(t - 1)}{t + 1}} \ge 2^3$ $\frac{t^2 - t - 6}{t + 1} \ge 0$

Step 7: Solve the Rational Inequality and Apply the Domain Restriction

Factoring the numerator, we have: $\frac{(t - 3)(t + 2)}{t + 1} \ge 0$ The critical points are $t = -2, -1, 3$. We test the intervals:

• $t < -2$: $\frac{(-)(-)}{(-)} = (-) < 0$
• $-2 \le t < -1$: $\frac{(-)(+)}{(-)} = (+) \ge 0$
• $-1 < t < 3$: $\frac{(-)(+)}{(+)} = (-) < 0$
• $t \ge 3$: $\frac{(+)(+)}{(+)} = (+) \ge 0$

So the solution is $t \in [-2, -1) \cup [3, \infty)$. Applying the domain restriction $t \ge 0$, we get $t \in [3, \infty)$. Explanation: This step solves the rational inequality using critical points and interval testing. It's crucial to then apply the domain restriction $t \ge 0$ to obtain the valid range for $t$.

$\frac{(t - 3)(t + 2)}{t + 1} \ge 0$ $t \in [3, \infty)$

Step 8: Determine the Least Value of |z|

Since $t = |z| \in [3, \infty)$, the least value of $|z|$ is 3. Explanation: The least value of $|z|$ is the smallest number in the interval $[3, \infty)$, which is 3.

$|z| = 3$

Common Mistakes & Tips

• Remember that $|z|$ is always non-negative. This restriction is essential for discarding extraneous solutions.
• Carefully handle the signs when solving rational inequalities. Interval testing is critical.
• Pay close attention to logarithm and exponential properties. A misunderstanding can lead to errors in simplification.

Summary

By substituting $|z|$ with $t$, simplifying the inequality using properties of logarithms and exponentials, solving the resulting rational inequality, and applying the domain restriction $t \ge 0$, we found that $|z| \in [3, \infty)$. Therefore, the least value of $|z|$ is 3.

Final Answer

The final answer is \boxed{3}, which corresponds to option (B).