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Question Number 156362 by MathSh last updated on 10/Oct/21
Answered by mindispower last updated on 10/Oct/21
∣Ω1+iΩ2∣2=Ω12+Ω22Ω1+iΩ2=∫−∞∞ei(e2x+e−2x)ex−sh2(x)=∫0∞ei(e2x+e−2x)e−sh2(x)(ex+e−x)dxletu=sh(t)=du=ch(t)dt=2∫0∞ei(4sh2(x)+1)e−u2du=2ei∫0∞e(−1+4i)u2du∫0∞eau2du=f(a),Re(a)<0f(a)=π2aΩ1+iΩ2=ei(−1+4i)12π∣Ω1+iΩ2∣=π17⇒Ω12+Ω22=π17
Commented by MathSh last updated on 10/Oct/21
PerfectmydearSer,thankyou
Commented by mindispower last updated on 11/Oct/21
withepleasur
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