number system~help
2011-11-27 4:35 am
Let a and b be non-zero real number. If the result of (a+bi)(-10+25i) is a real number, find a:b
回答 (3)
2011-11-27 6:28 am
✔ 最佳答案
(a+bi)(-10+25i)=-10a+25ai-10bi-25b=(-10a-25b)+(25a-10b)i Since (a+bi)(-10+25i) is real,so (25a-10b)i=0 25a-10b=0 a=(10b)/25 a/b=2/5 a:b=2:5
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2011-11-27 6:10 am
-10+25i=-5(-2+5i)
(a+bi)(-10+25i)
=-5(a+bi)(-2+5i)
=-5[(-2a-5b)+(5a-2b)i]
=(10a+25b)+(10b-25a)i
since (a+bi)(-10+25i)=(10a+25b)+(10b-25a)i is a real number,
imaginary part =0
10b-25a=0
10b=25a
2b=5a
a:b=2:5
(a+bi)(-10+25i)
=-5(a+bi)(-2+5i)
=-5[(-2a-5b)+(5a-2b)i]
=(10a+25b)+(10b-25a)i
since (a+bi)(-10+25i)=(10a+25b)+(10b-25a)i is a real number,
imaginary part =0
10b-25a=0
10b=25a
2b=5a
a:b=2:5
2011-11-27 5:11 am
原該我用中文答
(a+bi)(-10+25i) 係實數的話
(a+bi)一定係(-10+25i)的共軛複數
=>a=-10 b=-25
(a+bi)(-10+25i) 係實數的話
(a+bi)一定係(-10+25i)的共軛複數
=>a=-10 b=-25
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20111126000051KK00786