What's the value of x when sin⁡x=2 and x is a complex number?

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What&#039;s the value of x when sin⁡x=2 and x is a complex number?

peterpan7117i · Accepted Answer

I like these. I think I did sin⁡x=8 in a previous answer.  eix=cos⁡x+isin⁡x  eix=cos⁡x−isin⁡x  eix−e−ix=2isin⁡x  sin⁡x=eix−e−ix2i  Let y=eix. Then e−ix=1y.  2(2i)=y−1y  y2−4iy−1=0  y=2i±−3=i(2±3)  Convert to polar and get everything in the exponent. We note eiπ2=i and e2πki=1 for integer k.  eix=y=eiπ2eln⁡(2±3)e2πki  ix=iπ2+ln⁡(2±3)+2πki  x=π2+2πk−iln⁡(2±3) for integer k  Check: Let’s check one, plus root, k=0.  x=π2−iln⁡(2+3)  eix=eiπ2(2+3)=i(2−3)  e−ix=1i(2−3)=−i(2+3)=i(−2−3)  eix−e−ix=4i  sin⁡x=eix−e−ix2i=2

Esther Phillips · Accepted Answer

sin⁡x=2We use the fact thatsin⁡x=eix−e−ix2iSo our equation iseix−e−ix2i=2Now, set z=eix. So we havez−1z2i=2which can be rearranged asz2−4iz−1=0Now, solve this(z−2i)2+4−1=0in other words(z−2i)2−(i3)2=0which is also(z−(2+3)i)(z−(2−3)i)=0So we have two solutions:z1=(2+3)i and z2=(2−3)iNow, let’s get back to our initial variable x.eix=(2+3)iimplies that x=−iln⁡i−iln⁡(2+3)+2kπ,(k∈Z)=2kπ+π2−iln⁡(2+3)Likewise, we find that x′=2kπ+π2−iln⁡(2−3) are the other class of solutions.Hence, the solutions are (2k+12)π−iln⁡(2±3) where k∈Z.

What's the value of x when \sin x=2 and x

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