"Solving a trigonometric equation for purely imaginary numbers"By..." was answered by our community

Litzy Wallace

Answered question

2022-04-01

Solving a trigonometric equation for purely imaginary numbers
"By constraining z to be purely imaginary, show that the equation

\cos (z) = 2

can be represented as a standard quadratic equation. Solve this equation for z."

Jaslyn Allison

Beginner2022-04-02Added 13 answers

Step 1

\cos (x) = \frac{e^{i x} + e^{- i x}}{2}

So,

\cos (z) = \frac{e^{i z} + e^{- i z}}{2} = 2

(e^{i z} + e^{- i z}) = 4

e^{2 i z} + 1 = 4 e^{i z}

y^{2} - 4 y + 1 = 0

where

y = e^{i z}

Step 2
Which is a quadratic

y = (\frac{4 \pm \sqrt{16 - 4}}{2})

e^{i z} = (\frac{4 \pm \sqrt{16 - 4}}{2})

= 2 \pm \sqrt{3}

i z = \ln (2 \pm \sqrt{3})

z = - \ln (2 \pm \sqrt{3}) i

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