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OpenStudy (michele_laino):
hint:
if
z= a+ bi, then:
-z = -a - bi
OpenStudy (michele_laino):
furthermore:
z-6z=-5z
OpenStudy (chaylaceyx3):
sorry its actually the conjugate of z, not -z so its a-bi
OpenStudy (michele_laino):
ok!
then your equation can be rewritten as follows:
\[\Large z + 6\bar z = a + bi + 6a - 6bi = 7a - 5bi = 7\]
OpenStudy (michele_laino):
which is equivalent to these two real equations:
\[\Large \left\{ \begin{gathered}
7a = 7 \hfill \\
- 5b = 0 \hfill \\
\end{gathered} \right.\]
please solve that system for a and b
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OpenStudy (chaylaceyx3):
a=1 and b=0 ?
OpenStudy (michele_laino):
that's right!
so, what is z=...?
OpenStudy (chaylaceyx3):
do i just plug these numbers into 7a-5bi?
OpenStudy (michele_laino):
no, you have to plug, those values, for a and b, into z=a+bi
OpenStudy (michele_laino):
what do you get?
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OpenStudy (chaylaceyx3):
so its 1?
OpenStudy (michele_laino):
correct!
z=1 is the solution of your complex equation
OpenStudy (chaylaceyx3):
OH ok! Thanks so much!
OpenStudy (michele_laino):
:)
OpenStudy (chaylaceyx3):
question, where does the i go?
my teacher asked and i wasn't sure