How should the equation be solved&nbsp;cos⁡z=2&nbsp;for z?

Step 1 Euler's formula gives: cos⁡(z)=2⇔eiz+e−iz2=2 ⇔eiz+e−iz=22⇔(eiz)2−22×eiz+1=0 ⇔(eiz−2)2=1⇔eiz=2±1 ⇔iz+2ikπ=ln⁡(2±1) with k∈Z ⇔z=−iln⁡(2±1)+2lπ with l∈Z(l=−k)

Step 1 z=−iln⁡(2−1) Proof: exp⁡(iz)=cos⁡(z)+isin⁡(z)=2−1 This uses sin2⁡z=1−cos2⁡z=1−2=−1, so sin⁡(z)=i Hence iz=ln⁡(2−1), so z=−iln⁡(2−1)

Step 1&nbsp;cos⁡(z)=(eiz+e−iz2=2&nbsp;This equation may be revised&nbsp;ei2z−22eiz+1=0&nbsp;We make&nbsp;y=eiz&nbsp;y2−22y+1=0&nbsp;This quadratic equation has two roots, namely:&nbsp;y1=2+1; and&nbsp;y2=2−1&nbsp;Consequently, the answer is:&nbsp;z1=−iln⁡(2+1)=−i0.88137;&nbsp;z2=−iln⁡(2−1)=i0.88137

Step 1 Euler's formula gives: cos⁡(z)=2⇔eiz+e−iz2=2 ⇔eiz+e−iz=22⇔(eiz)2−22×eiz+1=0 ⇔(eiz−2)2=1⇔eiz=2±1 ⇔iz+2ikπ=ln⁡(2±1) with k∈Z ⇔z=−iln⁡(2±1)+2lπ with l∈Z(l=−k)

Step 1 z=−iln⁡(2−1) Proof: exp⁡(iz)=cos⁡(z)+isin⁡(z)=2−1 This uses sin2⁡z=1−cos2⁡z=1−2=−1, so sin⁡(z)=i Hence iz=ln⁡(2−1), so z=−iln⁡(2−1)

Step 1&nbsp;cos⁡(z)=(eiz+e−iz2=2&nbsp;This equation may be revised&nbsp;ei2z−22eiz+1=0&nbsp;We make&nbsp;y=eiz&nbsp;y2−22y+1=0&nbsp;This quadratic equation has two roots, namely:&nbsp;y1=2+1; and&nbsp;y2=2−1&nbsp;Consequently, the answer is:&nbsp;z1=−iln⁡(2+1)=−i0.88137;&nbsp;z2=−iln⁡(2−1)=i0.88137

Step by step solution to "How should the equation be solvedcos⁡z=2for z?"

Answered question

2022-01-17

How should the equation be solved $\cos z = \sqrt{2}$ for z?

Answer & Explanation

nick1337

Expert2022-01-17Added 777 answers

Step 1
Euler's formula gives:
$\cos (z) = \sqrt{2} \Leftrightarrow \frac{e^{i z} + e^{- i z}}{2} = \sqrt{2}$
$\Leftrightarrow e^{i z} + e^{- i z} = 2 \sqrt{2} \Leftrightarrow (e^{i z})^{2} - 2 \sqrt{2} \times e^{i z} + 1 = 0$
$\Leftrightarrow (e^{i z} - \sqrt{2})^{2} = 1 \Leftrightarrow e^{i z} = \sqrt{2} \pm 1$
$\Leftrightarrow i z + 2 i k π = \ln (\sqrt{2} \pm 1)$ with $k \in Z$
$\Leftrightarrow z = - i \ln (\sqrt{2} \pm 1) + 2 l π$ with $l \in Z (l = - k)$

star233

Skilled2022-01-17Added 403 answers

Step 1
$z = - i \ln (\sqrt{2} - 1)$
Proof:
$\exp (i z) = \cos (z) + i \sin (z) = \sqrt{2} - 1$
This uses $\sin^{2} z = 1 - \cos^{2} z = 1 - 2 = - 1$ , so $\sin (z) = i$
Hence $i z = \ln (\sqrt{2} - 1)$ , so $z = - i \ln (\sqrt{2} - 1)$

alenahelenash

Expert2022-01-24Added 556 answers

Step 1 $\cos (z) = (e^{i z} + \frac{e^{- i z}}{2} = \sqrt{2}$ This equation may be revised $e^{i 2 z} - 2 \sqrt{2} e^{i z} + 1 = 0$ We make $y = e^{i z}$ $y^{2} - 2 \sqrt{2} y + 1 = 0$ This quadratic equation has two roots, namely: $y_{1} = \sqrt{2} + 1$ ; and $y_{2} = \sqrt{2} - 1$ Consequently, the answer is: $z_{1} = - i \ln (\sqrt{2} + 1) = - i 0.88137$ ; $z_{2} = - i \ln (\sqrt{2} - 1) = i 0.88137$

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