情報制御数学複素解析レポート2回目回答

情報制御数学複素解析レポート２回目回答
平成 26 年 4 月 2 日
課題 1
z1 = r1 (cos θ1 + j sin θ1 )
z2 = r2 (cos θ2 + j sin θ2 )
r2 (cos θ2 + j sin θ2 )
z2
=
z1
r1 (cos θ1 + j sin θ1 )
=
r2 (cos θ2 + j sin θ2 )(cos θ1 − j sin θ1 )
r1 (cos θ1 + j sin θ1 )(cos θ1 − j sin θ1 )
=
r2 cos θ1 cos θ2 + sin θ1 sin θ2 + j(sin θ1 cos θ2 + cos θ1 sin θ2 )
r1
(cos2 θ1 + sin2 θ1 )
=
r2
(cos θ1 cos θ2 + sin θ1 sin θ2 + j(cos θ1 sin θ2 − sin θ1 cos θ2 )
r1
=
r2
(cos(θ2 − θ1 ) + j sin(θ2 − θ1 ))
r1
∴
r2
|z2 |
z2
=
=| |
r1
|z1 |
z1
arg(
z2
) = θ2 − θ1 = arg z2 − arg z1
z1
1
課題 2
f (z) = cos z =
ejz + e−jz
1
= (ej(x+jy) + e−j(x+jy) )
2
2
1
1
= (e(−y+jx) + e(y−jx) ) = (e−y (cos x + j sin x) + ey (cos x − j sin x))
2
2
1
= (e−y cos x + ey cos x + j(e−y sin x − ey sin x))
2
ux = −e−y sin x − ey sin x , vx = e−y cos x − ey cos x
uy = −e−y cos x + ey cos x , vy = −e−y sin x − ey sin x
1
f ′ (z) = (−e−y sin x − ey sin x + j(e−y cos x − ey cos x))
2
=
1
(−(e−y cos x − ey cos x) − j(e−y sin x + ey sin x))
2j
=
1 y
(e (cos x − j sin x) − e−y (cos x + j sin x))
2j
=
1 −jz
ejz − e−jz
(e − ejz ) = −
= − sin z
2j
2j
f (z) = sin z =
=
ejz − e−jz
2j
ejz e−jz
−
2j
2j
ejz
e−jz
f (z) = j
+j
2j
2j
′
=
ejz + e−jz
= cos z
2
2

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