Question 87511: This question is really troubling me and i could use some help please. Given that “e” is a complex number, solve e + 6ē = 7 (note: ē is the complex conjugate of e)
Found 2 solutions by stanbon, rapaljer:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Given that “e” is a complex number, solve e + 6ē = 7 (note: ē is the complex conjugate of e)
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If e is a complex number, e = a + bi
If e-bar is its conjugate, e-bar = a-bi
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Substituting:
a+bi + 6(a-bi) = 7
7a -5bi =7 + 0i
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7a = 7, so a = 1
-5b = 0, so b = 0
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So e = 1 + 0i
e = 1
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Cheers,
stan H.
Answer by rapaljer(4671)  (Show Source):
You can put this solution on YOUR website! Let e= a+bi, and e(complememt) = a-bi
What you have is
e + 6ē = 7
a+bi + 6(a-bi) = 7 +0i
a+bi +6a -6bi = 7 + 0i
7a -5bi = 6 + 0i
Therefore, 7a= 6, so a=6/7
and -5b= 0, so b=0
Therefore, e=6/7 +0i
R^2 Retired from SCC
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