複數,COMPLEX NUMBER 等待高手解答，等待只因渴求!

Please find my deductions below:

圖片參考：http://f7.yahoofs.com/users/458862c7mf9ddb2ca/f6cbscd/__sr_/1075scd.jpg?phAV0oFBF3KIkLLO

Now, let y = z – 1, then it simplifies to:

圖片參考：http://f7.yahoofs.com/users/458862c7mf9ddb2ca/f6cbscd/__sr_/e75escd.jpg?phAV0oFBSQDffYA1

Then denote y = a + bi where a and b are real values:

圖片參考：http://f7.yahoofs.com/users/458862c7mf9ddb2ca/f6cbscd/__sr_/7742scd.jpg?phAV0oFBOo36QRtc

Comparing the real and imaginary parts, we have:

圖片參考：http://f7.yahoofs.com/users/458862c7mf9ddb2ca/f6cbscd/__sr_/1554scd.jpg?phAV0oFBHi00wLQZ
.... (1)

圖片參考：http://f7.yahoofs.com/users/458862c7mf9ddb2ca/f6cbscd/__sr_/7459scd.jpg?phAV0oFBdITgqVWG
.... (2)
Sub (2) into (1), we have:

圖片參考：http://f7.yahoofs.com/users/458862c7mf9ddb2ca/f6cbscd/__sr_/2c03scd.jpg?phAV0oFBtKQaKCkP

After expanding the equation, we have:

圖片參考：http://f7.yahoofs.com/users/458862c7mf9ddb2ca/f6cbscd/__sr_/16a2scd.jpg?phAV0oFBjt.icLMx


which is a polynomial equation of b of degree 8.
By using numerical approximation methods for roots like Newton’s or bisection method, we can obtain one of the root for (*) is b = 2.3755 (corr. to 4 d.p.).
Sub this result into (2), we get a = 0.1581 (corr. to 4 d.p.).
Therefore, y = 0.1581 + 2.3755 i
With z – 1 = y, we have z = 1.1581 + 2.3755 i.
Direct substitution of z to the original equation can yield the result.
It is worth to note that according to (*), there may be at most 8 possible values of b (and hence a). So to speak, other 7 possible solutions may be obtained through solving for the remaining 7 roots of (*) (if exist).

2007-01-09 16:23:09 補充：
(*) is the polynomial equation of b of degree 8.