Complex Number-how to show it ?

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The discussion focuses on demonstrating that if w is totally imaginary for the complex number z = x + yi, then the equation x^2 + y^2 + 2x - 48 = 0 holds true. Participants express confusion about deriving this equation, with one individual successfully arriving at x^2 + y^2 - 6x + 8y = 0 but struggling to connect it to the required form. The importance of converting division by a complex number into multiplication is highlighted as a potential method for solving the problem. There is also a mention of verifying the conditions under which w remains totally imaginary. Ultimately, the conversation emphasizes the need for careful manipulation of complex numbers to reach the desired conclusion.
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Complex Number---how to show it ?

Given that z = x + yi and
w = ( z + 8i )/(z - 6) , z not equal to 6 .
If w is totally imaginary, show that :
x^2 + y^2 + 2x - 48 = 0

i understand the question, but the problem i facing is i only be able to show :
x^2 + y^2 - 6x + 8y = 0
i think that in order to satisfy what the question ask , i need to find y in term of x, but i can't do it...i don't sure whether is the question wrong already or my mistake. Any expert there, please help.
 
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w is totally imaginary <=> Re(w) = 0 <=> ...
 
I think there are too many continuations of that ellipsis (well, two), so as an aid, have you been taught how to convert division by a complex number into mulitplication by a complex number (one written as real plus i times imaginary)?

1/z = z*/(|z|^2)

now look at the imaginary part

(for muzza the other posibility i thought of involved the argument which didn't seem useful, though that was only a first impression)
 
scoutfai said:
Given that z = x + yi and
w = ( z + 8i )/(z - 6) , z not equal to 6 .
If w is totally imaginary, show that :
x^2 + y^2 + 2x - 48 = 0

If z=0 then w=-4i/3 which is totally imaginary, but -48<>0

I think the numerator should be (z + 8)
 
yes it should be (z+8)
simply note that (x+8)(x-6) = x^2+2x-48
 
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