The question:
Given that the expression (a + 2i)(6-2i) is a real number, find the value of real number a.
Solution:
(a+2i)(6-2i)
= 6a + 4 + (12 - 2a)i
Since (a+2i)(6-2i) is a real number, the imaginary part equals zero.
so 12 - 2a = 0
12 = 2a
a = 6
My question is
Why
"Since (a+2i)(6-2i) is a real number, the imaginary part equals zero."?
虛數的疑問,imaginary part = 0???
2009-08-04 5:49 am
回答 (2)
2009-08-04 6:42 am
✔ 最佳答案
if the imaginary part is not zero, it is not a real number but a complex number.for example:
1+2i
1+i
2i
and every real number has the imaginary part equlas zero, for example:
1=1 x 0i
2=2 x 0i
and most importantly:
0i=0
so, a number is a real number if and only if the imaginary part is zero.
2009-08-04 6:43 am
For any complex number z=a+bi , where i^2=-1
real number => b=0 , it's because real number does not include the imaginary i
purely imaginary number => a=0 , no real part is included.
real number => b=0 , it's because real number does not include the imaginary i
purely imaginary number => a=0 , no real part is included.
收錄日期: 2021-04-22 00:47:30
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090803000051KK02308