# Complex Number Problem

1. Jun 12, 2008

### crays

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?

Last edited: Jun 12, 2008
2. Jun 12, 2008

### tiny-tim

Hi crays!
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).

3. Jun 12, 2008

### crays

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 @_@.

4. Jun 12, 2008

### Defennder

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?

5. Jun 12, 2008

### tiny-tim

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 …

show us your working!

6. Jun 12, 2008

### crays

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

7. Jun 12, 2008

### HallsofIvy

Staff Emeritus
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?

8. Jun 12, 2008

### crays

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)

9. Jun 12, 2008

### crays

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 @_@

10. Jun 12, 2008

### Staff: Mentor

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

11. Jun 12, 2008

### tiny-tim

crays, the conjugate of z - 6 is not z +6, is it?

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

12. Jun 12, 2008

### Staff: Mentor

An expression is totally imaginary or pure imaginary if the real part is identically zero.

13. Jun 12, 2008

### crays

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 ?

14. Jun 12, 2008

### Staff: Mentor

Show your work, please.

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.

15. Jun 12, 2008

### tiny-tim

(x² - y² - 36) is wrong.

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

16. Jun 12, 2008

### crays

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)

17. Jun 12, 2008

### tiny-tim

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

Follow the hint!

18. Jun 12, 2008

### crays

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)

19. Jun 12, 2008

### tiny-tim

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?

20. Jun 12, 2008

### crays

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

Last edited: Jun 12, 2008
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