solve for z, and give your answer in the form a + bi (conj)z+2z = 2+4i+conj(7+3i) **conj refers to conjugate of a complex number (a line on top of it)

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solve for z, and give your answer in the form a + bi (conj)z+2z = 2+4i+conj(7+3i) **conj refers to conjugate of a complex number (a line on top of it)

Mathematics
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answer is 3+i (need to learn how to do it)
I don't understand the notation of the equation. Why did you put (conj)? Does it mean take the conjugate of the term that follows it? Or multiply that term by its conjugate? Or what?
i dont know how to write a line above the letter for this

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so for this question, there would be a line above the z and the 7+3i
z = a + bi (conj)z = a - bi
it means to flip the sign of the imaginary part
So why don't you just write z - 2z = 2 + 4i + (7 - 3i)? Is that what the question says?
no, there is no line above the z+2z, just the z
Can you draw the question instead?
|dw:1353720314063:dw|
\(\bar{z} + 2z = 2 + 4i + \overline{7+3i} = 2 + 4i + 7 - 3i = 9 + i\)
answer is 3+i though...you seem to be on the right track though
no idea what the conjugate of a variable means...
unless you use fourier which i don't think you are
It's not my problem statement. You need to be on the right track. Where shall we go from here? \(\bar{z} + 2z = 9 + i\)
i dont know what the z part is. that is the problem
is z considered real or imaginary?
i tried it with both for hours
Please read the problem statement. "solve for z, and give your answer in the form a + bi" z is a Complex Number
oh
You tried what? \(\bar{z} + 2z = a - bi + 2(a + bi) = a - bi + 2a + 2bi = (a + 2a) + (2b-b)i\) Do yuo see where this is going?
no
i understand why you get this but why did you do this
you equate this to 9+i?
That is it! \(3a + bi = 9 + i\)
thanks

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