F.4 數學 Number system 一條

If z is a complex number and (1+i)z^2－(2+3i)+2+i=0. Find z.請問如何計算？Sol(1+i)z^2－(2+3i)+2+i=0(1+i)z^2=2+3i－2－i(1+i)z^2=2iz^2=2i/(1+i)=2(1－i)i/2=i(1－i)=1+i=√2(1/√2+i/√2) =2^(1/2)(Cos(π/4)+iSin(π/4)) =2^(1/2)(Cos(2nπ+π/4)+iSin(2nπ+π/4)) z=2^(1/4)(Cos(nπ+π/8)+iSin(nπ+π/8)) (1) n=0z=2^(1/4) (Cos(π/8)+iSin(π/8)) =2^(1/4) (√(2+√2)/2+i√(2－√2)/2) =2^(－3/4) (√(2+√2)+i√(2－√2)) (2) n=1z=2^(1/4) (Cos(9π/8)+iSin(9π/8)) =2^(1/4) (－√(2+√2)/2－i√(2－√2)/2) =2^(－3/4) (－√(2+√2)－i√(2－√2)) 題目改為(1+i)z^2－(2+3i)z+2+i=0. (1+i)z^2－(2+3i)z+2+i=0D=(2+3i)^2－4*(1+i)(2+i) =(4+12i－9)－4(2+i+2i－1) =(4+12i－9)－4(1+3i) =－9(1+i)z^2－(2+3i)z+2+i=0z=((2+3i)+/-3i)/(2+2i) (1) z=(2+3i+3i)/(2+2i) =(1+3i)/(1+i) =(1+3i)(1－i)/[(1+i)(1－i)] =(1－i+3i+3)/2=2+i(2) z=((2+3i)－3i)/(2+2i) =2/(2+2i) =1/(1+i) =(1－i)/2

2010-09-03 00:15:19 補充：
If z is a complex number and (1+i)z^2－(2+3i)+2+i=0. Find z.
請問如何計算？
Sol
(1+i)z^2－(2+3i)+2+i=0
(1+i)z^2=2+3i－2－i
(1+i)z^2=2i
z^2=2i/(1+i)=2(1－i)i/2=i(1－i)=1+i
=√2(1/√2+i/√2)
=2^(1/2)(Cos(π/4)+iSin(π/4))
=2^(1/2)(Cos(2nπ+π/4)+iSin(2nπ+π/4))
z=2^(1/4)(Cos(nπ+π/8)+iSin(nπ+π/8))

2010-09-03 00:16:16 補充：
(1) n=0
z=2^(1/4) (Cos(π/8)+iSin(π/8))
=2^(1/4) (√(2+√2)/2+i√(2－√2)/2)
=2^(－3/4) (√(2+√2)+i√(2－√2))
(2) n=1
z=2^(1/4) (Cos(9π/8)+iSin(9π/8))
=2^(1/4) (－√(2+√2)/2－i√(2－√2)/2)
=2^(－3/4) (－√(2+√2)－i√(2－√2))

2010-09-03 00:16:53 補充：
題目改為(1+i)z^2－(2+3i)z+2+i=0.
(1+i)z^2－(2+3i)z+2+i=0
D=(2+3i)^2－4*(1+i)(2+i)
=(4+12i－9)－4(2+i+2i－1)
=(4+12i－9)－4(1+3i)
=－9

2010-09-03 00:17:01 補充：
(1+i)z^2－(2+3i)z+2+i=0
z=((2+3i)+/-3i)/(2+2i)
(1) z=(2+3i+3i)/(2+2i)
=(1+3i)/(1+i)
=(1+3i)(1－i)/[(1+i)(1－i)]
=(1－i+3i+3)/2
=2+i
(2) z=((2+3i)－3i)/(2+2i)
=2/(2+2i)
=1/(1+i)
=(1－i)/2

(1+i)z^2-(2+3i)+2+i=0
(1+i)z^2 -2 -3i +2 +i=0
(1+i)z^2 -3i +i = 0
(1+i)z^2 -2i = 0
(1+i)z^2 = +2i
z^2 = +2i/(1 + i)

         2i(1 – i)
z^2 = ---------------
        (1+i)(1 – i)

         2i(1 – i)
z^2 = ------------  (Note: i^2 = -1)
       (1– i^2)

         2i(1 – i)
z^2 = ------------
          1–(-1)

         2i(1 – i)
z^2 = ------------
             2

z^2 = i – i^2

z^2 = i – (-1)

z^2 = 1 + i
 
Taking square root on both sides

z = square root of (1 + i)  or (1 + i)^0.5

好明顯有問題

你話無打錯, 都get 唔到問乜
-(2+3i)+2+i
點解會咁寫?

麻煩你寫清楚d

2010-08-25 22:53:45 補充：
Z cannot be found.

Z can only be determined in terms of i

Have you missed z next to (2+3i)?

2010-08-25 17:56:01 補充：
If question isn't wrong, then you can solve for z simply by rearranging the real terms and imaginary terms, and take the square root of both side to get z,