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

Question

1628 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-05T11:10:52

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

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

#e^ (ix) = cos x + i sin x#

# e^((23pi)/8i) = ( cos((23pi)/8) + i sin((23pi)/8))#

#=> -0.9239 + i 0.3827, " II Quadrant"#

# e^((19pi)/6i) = cos((19pi)/6) + i sin((19pi)/6))#

#=> -0.866 - i 0.5, " III Quadrant"#

#e^((23pi)/8i) - e^((19pi)/6i) = -0.9239 + i 0.3827 + 0.866 + i 0.5#

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