SOLUTION: 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)

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)
------------
If e is a complex number, e = a + bi
If e-bar is its conjugate, e-bar = a-bi
------------------
Substituting:
a+bi + 6(a-bi) = 7
7a -5bi =7 + 0i
-------
7a = 7, so a = 1
-5b = 0, so b = 0
----------
So e = 1 + 0i
e = 1
===============
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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SOLUTION: This question is really troubling me and i could use some help please. Given that �e� is a complex number, solve e + 6&#275; = 7 (note: &#275; is the complex conjugate of e)

SOLUTION: 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)