Find the real part and the imaginary part of eiz as well as those of c

1.	Find the real part and the imaginary part of eiz as well as those of cos z. For what values of z is the real part of eiz equal to cosz?
Answer» e^iz = e^ix−y = e^−y (cos x + isin x) = u + iv, e^−iz = e^−ix+y = e^y(cos x − isin x). Thus cosz = (e^iz+ e^−iz)/2 = cos x cosh y − isin x sinh y. If u is equal to cos z, then e^−y cosx = cos x cosh y − isinx sinh y. The real part and the imaginary part of the equation above are: (e^−y− cosh y) cosx = 0 and sinx sinhy = 0. If y ≠ 0, these two equalities give sinx = cosx = 0, which is not possible. Both equalities are satisfied if y = 0. Thus the real part of e^iz is equal to cos z only if z is real.

Find the real part and the imaginary part of eiz as well as those of cos z. For what values of z is the real part of eiz equal to cosz?

Answer»

e^iz = e^ix−y = e^−y (cos x + isin x) = u + iv,

e^−iz = e^−ix+y = e^y(cos x − isin x).

Thus

cosz = (e^iz+ e^−iz)/2 = cos x cosh y − isin x sinh y.

If u is equal to cos z, then

e^−y cosx = cos x cosh y − isinx sinh y.

The real part and the imaginary part of the equation above are:

(e^−y− cosh y) cosx = 0 and sinx sinhy = 0.

If y ≠ 0, these two equalities give sinx = cosx = 0, which is not possible. Both equalities are satisfied if

y = 0.

Thus the real part of e^iz is equal to cos z only if z is real.

Discussion

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