Complex Trig Multiplication!

Algebra Level 3

cos ⁡ ( z ) sin ⁡ ( z ) = i \large \cos(z)\sin(z)=i

The principle values of complex number z z satisfying the equation above can be expressed as π a + b 2 ln ⁡ ( 2 ± c ) i d \dfrac{\pi}{a}+\dfrac{b}{2}\ln\left(2\pm\sqrt{c}\right)i^d , where a a , b b , c c , and d d are positive integers with d d being the smallest possible. Find b a − c d \dfrac{b}{a}-cd .


The answer is -14.5.

James Watson by James Watson , 16, United Kingdom

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1 solution

Given that

cos ⁡ z sin ⁡ z = i 1 2 sin ⁡ ( 2 z ) = i sin ⁡ ( 2 z ) = 2 i By Euler’s formula e 2 i z − e − 2 i z 2 i = 2 i e 2 i z − e − 2 i z = − 4 e 4 i z + 4 e 2 i z − 1 = 0 ⟹ e 2 i z = − 4 ± 16 + 4 2 e 2 i ( x + i y ) = ± 5 − 2 Ler z = x + i y ; x , y ∈ R e − 2 y ( cos ⁡ ( 2 x ) + i sin ⁡ ( 2 x ) ) = ± 5 − 2 \begin{aligned} \cos z \sin z & = i \\ \frac 12 \sin (2z) & = i \\ \sin (2z) & = 2i & \small \blue{\text{By Euler's formula}} \\ \frac {e^{2iz}-e^{-2iz}}{2i} & = 2i \\ e^{2iz} - e^{-2iz} & = -4 \\ e^{4iz} + 4e^{2iz} - 1 & = 0 \\ \implies e^{2iz} & = \frac {-4\pm \sqrt{16+4}}2 \\ e^{2i(x+iy)} & = \pm \sqrt 5 - 2 & \small \blue{\text{Ler }z = x + iy; \ x, y \in \mathbb R} \\ e^{-2y}(\cos (2x) + i \sin (2x)) & = \pm \sqrt 5 - 2 \end{aligned}

Equating the imaginary parts of both sides sin ⁡ ( 2 x ) = 0 ⟹ x = π 2 \sin (2x) = 0 \implies x = \dfrac \pi 2 for principal value. Equating the real parts:

e − 2 y cos ⁡ π = ± 5 − 2 − e − 2 y = ± 5 − 2 − 2 y = ln ⁡ ( ± 5 + 2 ) y = − ln ⁡ ( ± 5 + 2 ) 2 \begin{aligned} e^{-2y} \cos \pi & = \pm \sqrt 5 - 2 \\ - e^{-2y} & = \pm \sqrt 5 - 2 \\ -2y & = \ln (\pm \sqrt 5 + 2) \\ y & = - \frac {\ln (\pm \sqrt 5 + 2)}2 \end{aligned}

Therefore z = x + i y = π 2 − i 2 ln ⁡ ( ± 5 + 2 ) = π 2 + i 3 2 ln ⁡ ( 2 + ± 5 ) z = x + iy = \dfrac \pi 2 - \dfrac i2 \ln (\pm \sqrt 5 + 2) = \dfrac \pi 2 + \dfrac {i^3}2 \ln (2+\pm \sqrt 5) and b 1 − c d = 1 2 − ( 5 ) ( 3 ) = − 14.5 \dfrac b1 - cd = \dfrac 12 - (5)(3) = \boxed{-14.5}

2 Helpful 1 Interesting 2 Brilliant 0 Confused

@Chew-Seong Cheong sir I am impressed by seeing your interest in posting good solutions.
Few hours ago , I was struggling this problem and you posted a nice solution (i just upvoted)
Can you help me in this problem Thanks in advance.

A Former Brilliant Member - 11 months, 1 week ago

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