How do you evaluate e^( ( 23 pi)/8 i) - e^( ( 19 pi)/6 i)e23π8i−e19π6i using trigonometric functions?

1 Answer
sankarankalyanam
Aug 5, 2018

color(green)(e^((23pi)/8i) - e^((19pi)/6i) = -0.0579 + 0.8827 i, " IIQuadrant"e23π8i−e19π6i=−0.0579+0.8827i, IIQuadrant

Explanation:

Trigonometric form of e^ (ix)eix, using Euler's Equation, is given by

e^ (ix) = cos x + i sin xeix=cosx+isinx

e^((23pi)/8i) = ( cos((23pi)/8) + i sin((23pi)/8))e23π8i=(cos(23π8)+isin(23π8))

=> -0.9239 + i 0.3827, " II Quadrant"⇒−0.9239+i0.3827, II Quadrant

e^((19pi)/6i) = cos((19pi)/6) + i sin((19pi)/6))e19π6i=cos(19π6)+isin(19π6))

=> -0.866 - i 0.5, " III Quadrant"⇒−0.866−i0.5, III Quadrant

e^((23pi)/8i) - e^((19pi)/6i) = -0.9239 + i 0.3827 + 0.866 + i 0.5e23π8i−e19π6i=−0.9239+i0.3827+0.866+i0.5

color(green)(e^((23pi)/8i) - e^((19pi)/6i) = -0.0579 + 0.8827 i, " IIQuadrant"e23π8i−e19π6i=−0.0579+0.8827i, IIQuadrant

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