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sin-z-200-find-z-




Question Number 19419 by NEC last updated on 10/Aug/17
sin z=200    find z
sinz=200findz
Commented by NEC last updated on 10/Aug/17
yes. i know that.  if its a complex number i′ll like  to see it.Thanks
yes.iknowthat.ifitsacomplexnumberillliketoseeit.Thanks
Answered by sma3l2996 last updated on 10/Aug/17
sinz=((e^(iz) −e^(−iz) )/(2i))=200⇔e^(iz) −e^(−iz) =400i  ; z=a+ib  e^(ia) e^(−b) −e^(−ia) e^b =400i  e^(−b) (cosa+isin(a))−e^b (cosa−isina)=400i  cos(a)(e^(−b) −e^b )+isin(a)(e^(−b) +e^b )=400i  cos(a)=0   sin(a)(e^(−b) +e^b )=400  a=(π/2)+kπ  (−1)^k (e^(−b) +e^b )=400  e^(2b) +1=(−1)^k 400e^b   e^(2b) +(−1)^(k+1) 400e^b +1=0  e^(2b) +(−1)^(k+1) 2×200e^b +200^2 −200^2 +1=0  (e^b +(−1)^(k+1) 200)^2 =4×10^4 −1  e^b =(√(39999))+(−1)^k 200  b=ln((√(39999))+(−1)^k 200)
sinz=eizeiz2i=200eizeiz=400i;z=a+ibeiaebeiaeb=400ieb(cosa+isin(a))eb(cosaisina)=400icos(a)(ebeb)+isin(a)(eb+eb)=400icos(a)=0sin(a)(eb+eb)=400a=π2+kπ(1)k(eb+eb)=400e2b+1=(1)k400ebe2b+(1)k+1400eb+1=0e2b+(1)k+12×200eb+20022002+1=0(eb+(1)k+1200)2=4×1041eb=39999+(1)k200b=ln(39999+(1)k200)
Commented by NEC last updated on 11/Aug/17
wow! thanks    how is a=(π/2)+kπ    ?
wow!thankshowisa=π2+kπ?
Commented by sma3l2996 last updated on 11/Aug/17
because cos(a)=0
becausecos(a)=0
Commented by NEC last updated on 11/Aug/17
thanks.... its understood.
thanks.itsunderstood.

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