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Another Motivational Argument for the Expression: e^{iq}= cos q+ i sin qProblem: Show that e^{iq} = cos q+ i sin q, where i = . Solution: Let z = x + iy be any complex number. We know, from geometry, that z = x + iy = r(cos q + i sin q). In the previous article on
e^{ix}, we used the theory of differential equations to establish
the required identity. This time, we will use the natural logarithm function
ln(z) to establish that same identity. ln(z) = ln(x
+ iy) The first term in the third
line, ln(), involves
the real number ,
and so will concern us no further here. The second term, ln(cos q + i
sin q), involves a complex number whose magnitude is unity. u(q) = ln(cos q + i sin q) Then e^{u(q)} = e^{ln(cos q + i sin q)} = cos q + i sin q. Our problem thus reduces to showing that u(q) = iq .We notice immediately that e^{u(0)} = cos 0 + i sin 0 = 1 which gives us the identity u(0) = 1. We now differentiate the expression e^{u} = cos q + i sin q to obtain de^{u}
= e^{u} du = (- sin q + i cos q) dq We now have the derivative dq/du: dq/du = (cos
q + i sin q)/(- sin q + i cos q) Multiplying by unity in the form (cos q - i sin q)/(cos q - i sin q) allows us to simplify the right-hand side, giving du = i dq. Integrating, we acquire u = iq + C. But since we already know that u = 0 when q = 0, we have that the constant C = 0. Therefore u = iq, which establishes the required
identity. |
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