Complex Logarithm (Part 2)

Algebra Level 3

$\large 3 \left[\ln \left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) - \ln \left(\frac{\sqrt{3}}{2} - \frac{1}{2}i\right) \right] = A +Bi$

$A$ and $B$ are real numbers satisfying the equation above. Find the minimum value of $A^2 + B^2$ to 2 decimal places.

Notations:

$i = \sqrt{-1}$ denotes the imaginary unit.
$\ln (\cdot)$ denotes the natural logarithm.

The answer is 9.87.

1 solution

Chew-Seong Cheong
Oct 19, 2016

$\begin{aligned} z & = 3 \left[\ln \left(\color{#3D99F6}{\frac{\sqrt{3}}{2} + \frac{1}{2}i} \right) - \ln \left(\color{#3D99F6}{\frac{\sqrt{3}}{2} - \frac{1}{2}i} \right) \right] & \small \color{#3D99F6}{\text{Using Euler's formula}} \\ & = 3 \left[\ln \left( \color{#3D99F6}{e^{\frac \pi 6 i}} \right) - \ln \left(\color{#3D99F6}{e^{-\frac \pi 6 i}} \right) \right] \\ & = 3 \ln \left( \frac{e^{\frac \pi 6 i}}{e^{-\frac \pi 6 i}} \right) \\ & = 3 \ln \left( e^{\frac \pi 3 i} \right) \\ & = 3 \cdot \frac \pi 3 i \\ & = \pi i \end{aligned}$

$\implies A^2 + B^2 = 0^2 + \pi^2 \approx \boxed{9.87}$

Complex Logarithm (Part 2)

The answer is 9.87.

1 solution

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