cos + i sin = cos + i sin = cos 2π iθ

Transcription

1 Complx numbrs 1C cosθ + sinθ = cosθ + i sin θ 1 a cosθ + sin θ = cos12θ + i sin12θ b 4 c d f 2 a b c d π π π π cos + isin = cos + isin 8 i = + π π 8π 8π 2π 2π cos + isin = cos + isin = cos + isin 1 i = + 2π 2π 10π 10π cos + i sin = cos + i sin = cos 2π+ i sin 2π = π π π π π π cos isin = cos + isin = cos( ) + isin( ) = i ( cosθ+ isinθ) cosθ + isinθ = = cosθ+ isinθ = 2 4 cos 2θ + sin 2θ cosθ+ sinθ 7 14 cos 2θ + sin 2θ cosθ+ sinθ = = (cosθ+ isin θ) = 12 cos 4θ + sin 4θ cosθ+ sinθ = = = iθ ( cos 2θ + i sin 2θ) ( cosθ + i sinθ) 4 8 iθ iθ iθ cos2θ + sin 2θ cosθ+ sinθ 1 1 = = = 9 iθ = cos θ + sin θ cosθ+ sinθ cosθ+ isinθ iθ iθ cosθ + isinθ cosθ+ sinθ cosθ + sinθ = = = cosθ isinθ cos( θ) + isin( θ) cos θ+ isin θ iθ = 11iθ Parson Education Ltd Copying prmittd for purchasing institution only. This matrial is not copyright fr. 1

2 2 f iθ cosθ isinθ cos( θ) + isin( θ ) = = = ( cos2θ isin 2θ) ( cos( 2 θ) + isin( 2 θ) ) ( cos θ+ isin θ) iθ iθ = iθ a 4 7π 7π 28π 28π 28πi cos + isin cos + isin = = 24πi = 4π 4π 24π 24π cos sin 1 cos sin + i i = = 4πi 1 2πi 1 b Rcall th idntity sin(π + x) = sin(π x ) thn π 11π π π 9π 9π 9πi cos isin cos + isin cos + isin = = = = 2πi 1π π π π 2π 2π cos sin 7 cos sin cos sin + + i + i i = = πi 1 7πi 7 c 7 7 4π 2π 4π 4π 28π 28π 28πi cos isin cos + isin cos + isin = = = = 1πi 10π 4π 4π 4π 1π 1π cos sin cos sin cos sin + + i + i i = = 4πi 1 12πi Parson Education Ltd Copying prmittd for purchasing institution only. This matrial is not copyright fr. 2

3 4 a (1+ i) If z= 1+ i, thn r= = π θ = argz= tan = 1 4 π π So, 1+ i= 2 cos + i sin π π (1+ i) = 2 cos + i sin π π = ( 2) cos + isin π π = 4 2 cos + isin Thrfor, 1 1 = 4 2 i = 4 4i (1+ i) = 4 4i ( 2) = 2 = 4 2 Parson Education Ltd Copying prmittd for purchasing institution only. This matrial is not copyright fr.

4 4 b ( 2+ 2i) If z= 2+ i, thn 8 r= ( 2) + 2 = 4+ 4 = 8= 4 2 = 1 2 π π θ = argz= π tan = π = 2 π π So, 2+ 2i= cos + i sin 8 π π ( 2+ 2i) = cos + isin Thrfor, 8 24π 24π = () cos + i sin = (2)(1) (cos π+ i sin π) = 409(1+ i(0)) = ( 2+ 2i) = Parson Education Ltd Copying prmittd for purchasing institution only. This matrial is not copyright fr. 4

5 4 c (1 i) If z= 1 i, thn r= = π θ = argz= tan = 1 4 π π So, 1 i= 2 cos + i sin π π (1 i) = 2 cos + i sin Thrfor, π π = ( 2) cos + isin π π = 8 cos + isin = 8( 0+ i) = 8i (1 i) = 8i Parson Education Ltd Copying prmittd for purchasing institution only. This matrial is not copyright fr.

6 4 d (1 i) If z= 1 i, thn r= 1 + ( ) = 1+ = 4 = 2 1 π θ = argz= tan = 1 π π So, 1 i= 2 cos + isin π π (1 i) = 2 cos + isin Thrfor, π π = (2) cos + isin = 4(cos( 2π) + isin( 2π)) = 4(1+ i(0) = 4 (1 i) = 4 Parson Education Ltd Copying prmittd for purchasing institution only. This matrial is not copyright fr.

7 4 1 i 1 If z= i, thn r= + = + = = π θ = argz= tan = tan = 2 1 π π So, i= cos isin π π i = cos + isin Thrfor, 9 9π 9π = ( ) cos + isin π π = 81 cos + isin = 81 (0+ i) = 81 i 1 i = 81 i 9 9 Parson Education Ltd Copying prmittd for purchasing institution only. This matrial is not copyright fr. 7

8 4 f ( 2 2i) If z= 2 2i, thn r= ( 2 ) + ( 2) = = 1 = π π θ = argz = π+ tan = π+ = 2 π π So, 2 2i= 4 cos + isin π π ( 2 2i) = 4 cos + isin Thrfor, 2π 2π = 4 cos + isin π π = 1024 cos + isin 1 = 1024 i = 12 12i ( 2 2i) = 12 12i Parson Education Ltd Copying prmittd for purchasing institution only. This matrial is not copyright fr. 8

9 (+ i) If z= + i, thn r= + ( ) = 9+ = 12 = 4 = 2 1 π θ = argz= tan = π π So, + i= 2 cos + isin π π (+ i) = 2 cos + isin π π = (2 ) cos + i sin Thrfor, 1 = 2(9 ) + i 1 = i = i = i (+ i) = i π π w= 2 cos + isin π π 2π 2π 1 i = 2 cos + sin = 1 cos + sin = w i i = 8+ 8i Parson Education Ltd Copying prmittd for purchasing institution only. This matrial is not copyright fr. 9

10 7 π π z= cos isin z = + i = + i π π 18π 18π cos sin 27 cos sin π π = 27 cos + isin = 27i 8 a Lt z= 1+ i and w= 1 i first of all not that z = w = 1+ = 2 so it follows that w can writ z 2 i θ = and w 2 i ϕ π π = with tanθ = and tanϕ = so that θ = and ϕ = thn it follows that iπ 1+ i z 2 = = = π 1 2 i i w b W hav n i2π i2nπ 1+ i 2nπ 2nπ = = cos + isin 1 i This is ral prcisly whn 2 n π π = k for som intgr k so th smallst valu of n for which this 2π is ral is n= 2 and w hav cos = 1so this is not admissibl, th nxt smallst valu for 2 which this is ral is n= in which cas smallst valu. 9 W start by noting that w can writ iθ a+ bi= r cosθ+ isinθ = r a bi= r(cosθ isin θ) = r Thn it follows that iθ 1+ i 1 i n n n inθ n inθ n inθ inθ ( a+ bi) + ( a bi) = r + r = r ( + ) which is ral. n n = r cosnθ + isinnθ + cosnθ isinnθ = 2r cosnθ = cos 2π= 1 which is positiv hnc this is th Parson Education Ltd Copying prmittd for purchasing institution only. This matrial is not copyright fr. 10

11 Challng + Givn n Z, w hav: (r(cosθ + isin)) n = 1 1 = ( r( cosθ+ isin) ) n r n ( cosnθ + isinnθ) by d Moivr s thorm for positiv intgr xponnts. 1 cosnθ isinnθ = n r cosnθ + isinnθ cosnθ isinnθ cosnθ isinnθ cosnθ isinnθ = = n r n n r n n 2 n ( cos θ i sin θ) ( cos θ+ sin θ) n ( cos θ isin θ) ( cos( θ) isin( θ) ) = = + n r n n r n n Parson Education Ltd Copying prmittd for purchasing institution only. This matrial is not copyright fr. 11