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Another Motivational Argument for the Expression: eiq= cos q+ i sin q
Problem: Show that eiq = 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
eix, 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
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)
eu(q) = eln(cos q + i sin q) = cos q + i sin q.
Our problem thus reduces to showing that u(q) = iq .We notice immediately that
eu(0) = cos 0 + i sin 0 = 1
which gives us the identity
u(0) = 1.
We now differentiate the expression eu = cos q + i sin q to obtain
= eu 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
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