close
標題:
complex number shading
http://i987.photobucket.com/albums/ae355/delaynopemore/078ba4a6.jpg 又有問題唔識..... 我想問 b part的 shading係點做架? 我係 a part 成功搵到 l z - 5 l = 2 但b part 的marking scheme係直接出答案 唔係要代 testing點 先得架咩? 教下我!!!!
最佳解答:
Let z' is the conjuate of z (a) |(z - 1)/(z - 4)| = 2 |z - 1| = 2|z - 4| (z - 1)(z - 1)' = 4(z - 4)(z - 4)' z^2 - (z + z') + 1 = 4[z^2 - 4(z + z') + 16] 3z^2 - 15(z + z') + 63 = 0 z^2 - 5(z + z') + 21 = 0 (z - 5)(z - 5)' + 21 - 25 = 0 |z - 5| = 2 (b) Method 1: |(z - 1)/(z - 4)| < 2 z^2 - (z + z') + 1 < 4[z^2 - 4(z + z') + 16] 3z^2 - 15(z + z') + 63 > 0 z^2 - 5(z + z') + 21 > 0 (z - 5)(z - 5)' + 21 - 25 > 0 |z - 5| > 2 So, the shaded area is the outside region of the circle |z - 5| = 2 Method 2 : Sub. 6 into |(z - 1)/(z - 4)| = 5/2 = 2.5 > 2. Since 6 is in |z - 5| = 2, we conclude that the shaded area is the outside region of the circle |z - 5| = 2.
其他解答:
你 a part 都成功搵到 l z - 5 l = 2, 你只是唔知佢代表什麼. 其實, z = x + yi, 所以 l z - 5 l = 2 ==> |x + yi - 5| = 2 ==> (x - 5)^2 + y^2 = 4 這是一條圓形方程, 圓心是 (5, 0), 半徑是 2. 所以 l z - 5 l < 2 的範圍是一個圓形, 就是以 (5, 0) 為圓心, 半徑為 2. 不包括圓周. (因為題目用 "<")
complex number shading
此文章來自奇摩知識+如有不便請留言告知
發問:http://i987.photobucket.com/albums/ae355/delaynopemore/078ba4a6.jpg 又有問題唔識..... 我想問 b part的 shading係點做架? 我係 a part 成功搵到 l z - 5 l = 2 但b part 的marking scheme係直接出答案 唔係要代 testing點 先得架咩? 教下我!!!!
最佳解答:
Let z' is the conjuate of z (a) |(z - 1)/(z - 4)| = 2 |z - 1| = 2|z - 4| (z - 1)(z - 1)' = 4(z - 4)(z - 4)' z^2 - (z + z') + 1 = 4[z^2 - 4(z + z') + 16] 3z^2 - 15(z + z') + 63 = 0 z^2 - 5(z + z') + 21 = 0 (z - 5)(z - 5)' + 21 - 25 = 0 |z - 5| = 2 (b) Method 1: |(z - 1)/(z - 4)| < 2 z^2 - (z + z') + 1 < 4[z^2 - 4(z + z') + 16] 3z^2 - 15(z + z') + 63 > 0 z^2 - 5(z + z') + 21 > 0 (z - 5)(z - 5)' + 21 - 25 > 0 |z - 5| > 2 So, the shaded area is the outside region of the circle |z - 5| = 2 Method 2 : Sub. 6 into |(z - 1)/(z - 4)| = 5/2 = 2.5 > 2. Since 6 is in |z - 5| = 2, we conclude that the shaded area is the outside region of the circle |z - 5| = 2.
其他解答:
你 a part 都成功搵到 l z - 5 l = 2, 你只是唔知佢代表什麼. 其實, z = x + yi, 所以 l z - 5 l = 2 ==> |x + yi - 5| = 2 ==> (x - 5)^2 + y^2 = 4 這是一條圓形方程, 圓心是 (5, 0), 半徑是 2. 所以 l z - 5 l < 2 的範圍是一個圓形, 就是以 (5, 0) 為圓心, 半徑為 2. 不包括圓周. (因為題目用 "<")
文章標籤
全站熱搜
留言列表
