# Complex imaginary Number Problem

Hi, i have this question which is related to complex number and i have just no idea how i should solve it. Some guide and help please.

Given that z = x + yi and w = (z+8i)/(z-6) , z $$\neq$$ 6. If w is totally imaginary, show that x^2 + y^2 + 2x - 48 = 0

I've tried alot of way comparing them. Just can't work.
I substituted z into w but end up still with a w. How can i get rid of the w?

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tiny-tim
Homework Helper
Hi crays!
I substituted z into w but end up still with a w.
You should really have shown us what you got, otherwise we can't see where you went wrong.

Hint: multiply top and bottom by the complex conjugate of (z - 6).

i tried using the conjugate way. The equation formed is extremely long @_@.
Here is it:

(x^2 - y^2 + 2x + 14yi + 48i) / (x^2 - y^2 - 36)

There is still i in it @_@.

Defennder
Homework Helper
You could just substitute the z into the expression for w, multiply the denominator with its complex conjugate. You are given that w is imaginary, so what does that say about that expression you have?

Anyway, I didn't manage to get that answer you gave. Did you transcribe the question correctly?

tiny-tim
Homework Helper
(x^2 - y^2 + 2x + 14yi + 48i) / (x^2 - y^2 - 36)

There is still i in it @_@.
crays, there's supposed t be an i in it!

w has to be purely imaginary … that's zero plus something-times-i.

If your equation were correct, the solution would be x² - y² + 2x = 0.

You have a minus wrong, and the 48 seems to be in the wrong place …

w = imaginary means that the whole w equation = imaginary ? I don't really understand how does that helps tho. Erm yes, i'll repeat my question here again

Given that z = x + yi and w = (z+8i)/(z-6) , z =/= 6 , show that x^2 + y^2 + 2x - 48 = 0

HallsofIvy
Homework Helper
Hi, i have this question which is related to complex number and i have just no idea how i should solve it. Some guide and help please.

Given that z = x + yi and w = (z+8i)/(z-6) , z $$\neq$$ 6. If w is totally imaginary, show that x^2 + y^2 + 2x - 48 = 0

I've tried alot of way comparing them. Just can't work.
I substituted z into w but end up still with a w. How can i get rid of the w?
You don't want to get rid of the w, you want to use the fact that it is "totally imaginary"-i.e., its real part is 0. If you put z= x+ yi into w, what is the real part of 0?

Okay, my working :
(z + 8i) / (z - 6) x (z + 6) / (z + 6)
= (z² - 6z + 8zi + 48i) / (z² - 36)
= [(x+yi)² - 6(x+yi) + 8(x+yi)i + 48i] / (x+yi)² - 36)
= (x² - y² - 6x + 6yi + 8x + 8yi + 48i) / (x² - y² - 36)
= (x² - y² + 2x + 14yi + 48i) / (x² - y² - 36)

You don't want to get rid of the w, you want to use the fact that it is "totally imaginary"-i.e., its real part is 0. If you put z= x+ yi into w, what is the real part of 0?
My teacher did not explain it properly. What does TOTALLY imaginary means? Means the whole equation is imaginary ? means there is no real number there? I'm confused @_@

D H
Staff Emeritus
Neither z+6 nor z-6 is not the complex conjugate of z-6. Your denominator is still complex.

tiny-tim
Homework Helper
(z + 8i) / (z - 6) x (z + 6) / (z + 6)
crays, the conjugate of z - 6 is not z +6, is it?

Hint: put z - 6 = x - 6 + yi.

D H
Staff Emeritus
An expression is totally imaginary or pure imaginary if the real part is identically zero.

So sorry my maths foundation and expansion is still not good enough. Here is my new one, (hopefully without error)

(x² - y² + 2x + 2yi - 48i) / (x² - y² - 36)

D H @ I don't really get it, so 0 is imaginary ?

D H
Staff Emeritus
(x² - y² + 2x + 2yi - 48i) / (x² - y² - 36)

D H @ I don't really get it, so 0 is imaginary ?
The same question (is zero real or imaginary) can be applied to the real numbers: is zero positive or negative? Zero can be viewed as both positive and negative, or as neither positive nor negative. The same applies to complex numbers. Zero is a special case.

tiny-tim
Homework Helper
(x² - y² - 36) is wrong.

Hint: what is (x - 6 - yi)(x - 6 + yi) ?

Sorry, here is my working.

[ (Z + 8i) / ( z - 6 ) ] x [ (z - 6) / (z - 6) ]
= (z² - 6z + 8zi - 48i) / (z² - 36)
= (x² - y² - 6x - 6yi + 8x + 8yi - 48i) / ( x² - y² - 36 )
= (x² - y² + 2x + 2yi - 48i) / (x² - y² - 36)

tiny-tim
Homework Helper
crays, both D H and I have told you that z+6 is not the complex conjugate of z-6.

i am, just that i posted before you posted lol. Thanks.
here it is, following ur hints, i've got.

(x² - y² + 2x + 2yi - 48i) / (x² - 12x + 2xyi + 36 - 12yi - y²)

tiny-tim
Homework Helper
i am, just that i posted before you posted lol. Thanks.
here it is, following ur hints, i've got.

(x² - y² + 2x + 2yi - 48i) / (x² - 12x + 2xyi + 36 - 12yi - y²)
Nooo. Let's concentrate on the denominator …

it should be purely real, shouldn't it?

The x² - 12x + 36 is correct, but the y part shouldn't have any i, should it?

finally!!! thanks.

I've got it cause in z there is no imaginary number so i must expand it first right?
with that i've found (denominator)
x² - 12x + 36 + y²

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tiny-tim
Homework Helper
finally!!! thanks.

I've got it cause in z there is no imaginary number so i must expand it first right?
with that i've found (denominator)
x² - 12x + 36 + y²
crays, before we go any further, can you please check the question …

is w = (z+8i)/(z-6), or is w = (z+8)/(z-6) ?

(I only get the right answer with the latter. )

Yes, the book says w = (z + 8i) / (z-6). Apparently my friend tried the question too but we both got stucked with (x² - y² -6x +8y -6yi + 8xi -48i)/(x² -12x + 36 + y²)

tiny-tim