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We have that cosh ( 2 ) + i sin ( x ) = e 2 1 . If we rewrite e 2 1 = e − 2 as cos θ + i sin θ , we see that e − 2 = e i ( 2 i ) . From this we have θ = 2 i
cos ( 2 i ) = 2 e i ( 2 i ) + e − i ( 2 i ) = 2 e − 2 + e 2 = cosh ( 2 ) ∴ cosh ( 2 ) + i sin ( 2 i ) = cos ( 2 i ) + i sin ( 2 i )
Hence, we see that x = 2 i ⟹ x = − 4
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cosh 2 + i sin x 2 e 2 + e − 2 + i ⋅ 2 i ( e − x i − e x i ) e 2 + e − 2 − e − x i + e x i e x i − e − x i ⟹ x i − x ⟹ x = e 2 1 = e − 2 = 2 e − 2 = e − 2 − e 2 = − 2 = 4 = − 4 Multiplying both sides by 2 Squaring both sides Note that i 2 = − 1