Using Euler's formula

Algebra Level 3

ln ( 3 2 + 1 2 i ) × ln ( 2 2 + 2 2 i ) × ln ( 1 2 + 3 2 i ) × ln ( i ) = π 4 k 2 \ln(\frac{\sqrt{3}}{2}+\frac{1}{2}i) \times \ln(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i) \times \ln(\frac{1}{2}+\frac{\sqrt{3}}{2}i) \times \ln(i) = \large \frac{\pi^{4}}{k^{2}}

Find the value of k |k| .

(The radian range: 0 x π 0 \leq x \leq \pi )

Hints: e i ( π ) = cos ( π ) + i sin ( π ) = 1 e^{i(\pi)} = \cos(\pi)+i\sin(\pi) = -1 \Rightarrow ln ( 1 ) = i π \ln(-1)=i\pi

Reference : Euler's Formula


The answer is 12.

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1 solution

Tommy Li
Jun 9, 2016

e i π 6 = cos ( π 6 ) + i sin ( π 6 ) = ( 3 2 + 1 2 i ) e^{i\frac{\pi}{6}} = \cos(\frac{\pi}{6})+i\sin(\frac{\pi}{6}) = (\frac{\sqrt{3}}{2}+\frac{1}{2}i)

e i π 4 = cos ( π 4 ) + i sin ( π 4 ) = ( 2 2 + 2 2 i ) e^{i\frac{\pi}{4}} = \cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4}) = (\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)

e i π 3 = cos ( π 3 ) + i sin ( π 3 ) = ( 1 2 + 3 2 i ) e^{i\frac{\pi}{3}} = \cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3}) = (\frac{1}{2}+\frac{\sqrt{3}}{2}i)

e i π 2 = cos ( π 2 ) + i sin ( π 2 ) = i e^{i\frac{\pi}{2}} = \cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2}) = i

ln ( 3 2 + 1 2 i ) × ln ( 2 2 + 2 2 i ) × ln ( 1 2 + 3 2 i ) × ln ( i ) = i π 6 × i π 4 × i π 3 × i π 2 = i 4 π 4 144 = π 4 1 2 2 \ln(\frac{\sqrt{3}}{2}+\frac{1}{2}i) \times \ln(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i) \times \ln(\frac{1}{2}+\frac{\sqrt{3}}{2}i) \times \ln(i) = i\frac{\pi}{6} \times i\frac{\pi}{4} \times i\frac{\pi}{3} \times i\frac{\pi}{2} = i^{4}\frac{\pi^{4}}{144}=\frac{\pi^{4}}{12^{2}}

k = 12 k = 12

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