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µoªí ¥Ñ benice ©ó ¬P´Á¤» ¤Q¤@¤ë 19, 2016 11:48 am


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¥O z = x + yi, x, y ¬°¹ê¼Æ¡C

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(x + yi)² - 2(1 + i)(x + yi) - 5 - 10i = 0
(x² - y² + 2xyi) - 2[x - y + (x + y)i] - 5 - 10i = 0
(x² - y² - 2x + 2y - 5) + 2(xy - x - y - 5)i = 0

©Ò¥H
ùú
¢xx² - y² - 2x + 2y - 5 = 0
¢xxy - x - y - 5 = 0    (½Ð°Ñ¦Ò¤U­±ªº¹Ï§Î)
ùü

°t¤è±o
ùú
¢x(x² - 2x + 1) - (y² - 2y + 1) - 5 = 0
¢x(x - 1)(y - 1) - 6 = 0
ùü
ùú
¢x(x - 1)² - (y - 1)² = 5
¢x(x - 1)(y - 1) = 6
ùü

¥O
a = x - 1
b = y - 1
±o
ùú
¢xa² - b² = 5 ...... (1)
¢xab = 6 ...... (2)
ùü

¥Ñ (2) ±o
b = 6/a ...... (3)

±N (3) ¥N¤J (1) ±o
a² - (6/a)² = 5
(a²)² - 5a² - 36 = 0
(a² - 9)(a² + 4) = 0
a² = 9
a = ¡Ó3

±N a = ¡Ó3 ¥N¤J (3) ±o
b = ¡Ó2

©Ò¥H
x = a + 1 = (3 + 1) ©Î (-3 +1) = 4 ©Î -2
y = b + 1 = (2 + 1) ©Î (-2 +1) = 3 ©Î -1

¦]¦¹¡A­ì¤èµ{¦¡ªº¸Ñ¬° 4 + 3i  ©Î  -2 - i¡C ¡½


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¥ªÁä: ÂIÀ»ÁY©ñ; ¥kÁä: Æ[¬Ý­ì¹Ï

µoªí ¥Ñ benice ©ó ¬P´Á¤» ¤Q¤@¤ë 19, 2016 9:51 am


z² - 2(1 + i)z - 5 - 10i = 0
z² - 2(1 + i)z - (5 + 10i) = 0

¥Ñ¤G¦¸¤èµ{¦¡ªº¤½¦¡¸Ñ¡A±o
z
= {2(1 + i) ¡Ó ¡Ô[4(1 + i)² + 4(5 + 10i)]} / 2
= (1 + i) ¡Ó ¡Ô[(1 + i)² + (5 + 10i)]
= (1 + i) ¡Ó ¡Ô[2i + 5 + 10i]
= (1 + i) ¡Ó ¡Ô[5 + 12i]
= (1 + i) ¡Ó ¡Ô[3² + 2*3*2i + (2i)²]
= (1 + i) ¡Ó ¡Ô[(3 + 2i)²]
= (1 + i) ¡Ó (3 + 2i)
= (1 + i) + (3 + 2i)  ©Î  (1 + i) - (3 + 2i)
= 4 + 3i  ©Î  -2 - i ¡½



©ÎªÌª½±µ¥Î°t¤èªk¡G
z² - 2(1 + i)z - 5 - 10i = 0
z² - 2(1 + i)z + (1 + i)² = 5 + 10i + (1 + i)²
[z - (1 + i)]² = 5 + 10i + 2i
[z - (1 + i)]² = 5 + 12i
[z - (1 + i)]² = 3² + 2*3*2i + (2i)²
[z - (1 + i)]² = (3 + 2i)²
z - (1 + i) = ¡Ó(3 + 2i)
z = 1 + i ¡Ó (3 + 2i) = 4 + 3i  ©Î  -2 - i ¡½

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µoªí ¥Ñ ³X«È ©ó ¬P´Á¤­ ¤Q¤@¤ë 18, 2016 10:32 pm

¨D«Y¼Æ½Æ¤èµ{¦¡z^2-2(1+I)z-5-10i=0 z¸Ñ¬°¡H
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