How many solutions does sin⁡(z)=10 have in the complex plane?

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Kirsten Davis · Accepted Answer

sin⁡is&amp;nbsp;2π- periodic, which means if you have a solution&amp;nbsp;z∈C&amp;nbsp;such as&amp;nbsp;sin⁡z=10, for any&amp;nbsp;k∈Z,sin⁡(z+2kπ)=10.&amp;nbsp;Also, for any&amp;nbsp;x∈R:&amp;nbsp;sin⁡(ix+π2)=ei(ix+π2)−e−i(ix+π2)2i&amp;nbsp;=e−xeiπ2−exeiπ22i&amp;nbsp;=ie−x+iex2i&amp;nbsp;=ex+e−x2=cos⁡h&amp;nbsp;x&amp;nbsp;We know that the range of&amp;nbsp;cos⁡h&amp;nbsp;is&amp;nbsp;[1,+∞), which implies there is at least one possible solution&amp;nbsp;x∈R&amp;nbsp;such as&amp;nbsp;cos⁡h&amp;nbsp;x=10&amp;nbsp;and at least one&amp;nbsp;z=ix+π2∈C&amp;nbsp;such as&amp;nbsp;sin⁡z=10.&amp;nbsp;As a result, the equation has an infinite number of solutions&amp;nbsp;sin⁡z=10∈C.

Janet Young · Accepted Answer

Just like with sin⁡x=12 there will be two supplementary angles that are solutions, that is, a pair of solutions that adds up to π. Then to each of those we can add 2πk for integer k to handle the periodicity. That’s a countably infinite number of solutions.These aren’t too hard to solve exactly. I’ll solve sin⁡z=a for real a,|a|&amp;gt;1The expression for sine from Euler’s Formula is:sin⁡z=12i(eiz−e−iz)Let&#039;s let x=eiz, then this becomesa=12i(x−1x)x2−2aix−1=0x=ai±−a2+1Since |a|&amp;gt;1 the radicand is negative, so x is purely imaginary.x=ai±−(a2−1)=i(a±a2−1)=eiπ2(a±a2−1)ln⁡x=izz=−iln⁡x=−i(iπ2+2πki+ln⁡(a±a2−1))z=(12+2k)π+iln⁡(a±a2−1)The cool thing about this is that the two expressions ln⁡(a±a2−1) are negations of each other, as the arguments to the log are reciprocals. (I’ll leave that to you to check. It follows from the product of the x is being -1, because of the -1 in the quadratic equation.) So the z s come in conjugate pairs.z=(12+2k)π±iln⁡(a+a2−1)Considering k=0 for a moment, the real part of both conjugates is π2, so when they’re summed they add to π as the imaginary parts cancel. They’re supplementary angles.Here we have a=10, soarcsin⁡10=(12+2k)π±iln⁡(10+99)So there are countably infinite solutions to sin⁡z=10, a conjugate pair of two solutions for each integer k.

How many solutions does \sin(z)=10 have in the complex plane?

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