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z= e^(1-i3) = e.e^(-i3)we know that e^-ix = cos x - i sin xso .pdf

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z= e^(1-i/3) = e.e^(-i/3) we know that e^-ix = cos x - i sin x so e^(-i/3) = cos /3 - i sin/3 = 1/2 - i3 /2 real part of e.e^(-i/3) = e/2 imaginary part = - e3 /2 Solution z= e^(1-i/3) = e.e^(-i/3) we know that e^-ix = cos x - i sin x so e^(-i/3) = cos /3 - i sin/3 = 1/2 - i3 /2 real part of e.e^(-i/3) = e/2 imaginary part = - e3 /2.

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z= e^(1-i/3) = e.e^(-i/3) we know that e^-ix = cos x - i sin x so e^(-i/3) = cos /3 - i sin/3 = 1/2 - i3 /2 real part of e.e^(-i/3) = e/2 imaginary part = - e3 /2 Solution z= e^(1-i/3) = e.e^(-i/3) we know that e^-ix = cos x - i sin x so e^(-i/3) = cos /3 - i sin/3 = 1/2 - i3 /2 real part of e.e^(-i/3) = e/2 imaginary part = - e3 /2

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z= e^(1-i3) = e.e^(-i3)we know that e^-ix = cos x - i sin xso .pdf

  • 1. z= e^(1-i/3) = e.e^(-i/3) we know that e^-ix = cos x - i sin x so e^(-i/3) = cos /3 - i sin/3 = 1/2 - i3 /2 real part of e.e^(-i/3) = e/2 imaginary part = - e3 /2 Solution z= e^(1-i/3) = e.e^(-i/3) we know that e^-ix = cos x - i sin x so e^(-i/3) = cos /3 - i sin/3 = 1/2 - i3 /2 real part of e.e^(-i/3) = e/2 imaginary part = - e3 /2