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3. We can extend the sine function to the complex numbers by defining \[ \sin ...
Mar 16, 2024
3. We can extend the sine function to the complex numbers by defining sin⁡(z):=eiz−e−iz2i for z∈C. \sin (z):=\frac{e^{i z}-e^{-i z}}{2 i} \quad \text { for } z \in \mathbb{C} . (a) Let α∈R\alpha \in \mathbb{R} with \alpha>1. Use the quadratic formula to show that if sin⁡(z)=−αi\sin (z)=-\alpha i, then we have eiz=α±α2+1. e^{i z}=\alpha \pm \sqrt{\alpha^{2}+1} . (b) Find all the solutions z∈Cz \in \mathbb{C} to the equation sin⁡(z)=−αi\sin (z)=-\alpha i, giving your solutions in Cartesian form.
Solution by Steps
step 1
Given the definition of the complex sine function: sin⁡(z)=eiz−e−iz2i\sin(z) = \frac{e^{iz} - e^{-iz}}{2i}
step 2
To find the values of zz such that sin⁡(z)=−αi\sin(z) = -\alpha i, we set the equation equal to −αi-\alpha i: eiz−e−iz2i=−αi\frac{e^{iz} - e^{-iz}}{2i} = -\alpha i
step 3
Multiplying both sides by 2i2i to eliminate the denominator, we get: eiz−e−iz=2αi2e^{iz} - e^{-iz} = 2\alpha i^2
step 4
Since i2=−1i^2 = -1, the equation simplifies to: eiz−e−iz=−2αe^{iz} - e^{-iz} = -2\alpha
step 5
Adding e−ize^{-iz} to both sides gives us: eiz=−2α+e−ize^{iz} = -2\alpha + e^{-iz}
step 6
Multiplying both sides by eize^{iz} to get a quadratic equation in terms of eize^{iz}: (eiz)2+2αeiz−1=0(e^{iz})^2 + 2\alpha e^{iz} - 1 = 0
step 7
Applying the quadratic formula to solve for eize^{iz}: eiz=−2α±(2α)2−4(−1)2e^{iz} = \frac{-2\alpha \pm \sqrt{(2\alpha)^2 - 4(-1)}}{2}
step 8
Simplifying under the square root and the denominator: eiz=−2α±4α2+42e^{iz} = \frac{-2\alpha \pm \sqrt{4\alpha^2 + 4}}{2}
step 9
Further simplification gives us: eiz=−α±α2+1e^{iz} = -\alpha \pm \sqrt{\alpha^2 + 1}
Answer
eiz=−α±α2+1e^{iz} = -\alpha \pm \sqrt{\alpha^2 + 1}
Key Concept
Quadratic Formula Application in Complex Analysis
Explanation
We used the definition of the complex sine function and set it equal to −αi-\alpha i. After simplifying and rearranging, we applied the quadratic formula to find the expression for eize^{iz} in terms of α\alpha.
Solution by Steps
step 1
From the previous result, we have eiz=−α±α2+1e^{iz} = -\alpha \pm \sqrt{\alpha^2 + 1}
step 2
To find all solutions z∈Cz \in \mathbb{C}, we use the logarithmic form for complex numbers: z=ln⁡(eiz)z = \ln(e^{iz})
step 3
Applying the logarithm to both sides of the equation from step 1: z=ln⁡(−α±α2+1)z = \ln(-\alpha \pm \sqrt{\alpha^2 + 1})
step 4
Since \alpha > 1, both α\alpha and α2+1\sqrt{\alpha^2 + 1} are real and positive, and the expression inside the logarithm is a complex number
step 5
The general solution for zz in terms of n∈Zn \in \mathbb{Z}, using the periodicity of the exponential function, is: z=ln⁡(−α±α2+1)+2πniz = \ln(-\alpha \pm \sqrt{\alpha^2 + 1}) + 2\pi ni
step 6
Expressing zz in Cartesian form, where z=x+yiz = x + yi, we have: x=ℜ(ln⁡(−α±α2+1))x = \Re(\ln(-\alpha \pm \sqrt{\alpha^2 + 1})) y=ℑ(ln⁡(−α±α2+1))+2πny = \Im(\ln(-\alpha \pm \sqrt{\alpha^2 + 1})) + 2\pi n
Answer
z=ln⁡(−α±α2+1)+2πniz = \ln(-\alpha \pm \sqrt{\alpha^2 + 1}) + 2\pi ni
Key Concept
Complex Logarithm and Periodicity
Explanation
We found the complex solutions for zz by taking the natural logarithm of the expression for eize^{iz} and considering the periodic nature of the complex exponential function, which introduces the 2πni2\pi ni term for n∈Zn \in \mathbb{Z}.
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