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

widdonod1t

widdonod1t

Answered

2022-01-16

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

Answer & Explanation

Kirsten Davis

Kirsten Davis

Expert

2022-01-17Added 27 answers

sinis 2π- periodic, which means if you have a solution zC such as sinz=10, for any kZ,sin(z+2kπ)=10
Also, for any xR
sin(ix+π2)=ei(ix+π2)ei(ix+π2)2i 
=exeiπ2exeiπ22i 
=iex+iex2i 
=ex+ex2=cosh x 
We know that the range of cosh is [1,+), which implies there is at least one possible solution xR such as cosh x=10 and at least one z=ix+π2C such as sinz=10
As a result, the equation has an infinite number of solutions sinz=10C.

Janet Young

Janet Young

Expert

2022-01-18Added 32 answers

Just like with sinx=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 sinz=a for real a,|a|>1
The expression for sine from Euler’s Formula is:
sinz=12i(eizeiz)
Let's let x=eiz, then this becomes
a=12i(x1x)
x22aix1=0
x=ai±a2+1
Since |a|>1 the radicand is negative, so x is purely imaginary.
x=ai±(a21)=i(a±a21)=eiπ2(a±a21)
lnx=iz
z=ilnx=i(iπ2+2πki+ln(a±a21))
z=(12+2k)π+iln(a±a21)
The cool thing about this is that the two expressions ln(a±a21) 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+a21)
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, so
arcsin10=(12+2k)π±iln(10+99)
So there are countably infinite solutions to sinz=10, a conjugate pair of two solutions for each integer k.

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