# Complex numbers : quadratic equation

1. May 26, 2016

### chwala

1. The problem statement, all variables and given/known data
Showing all necessary working solve the equation $iz^2+2z-3i=0$ giving your answer in the form $x+iy$ where x and y are real and exact,

2. Relevant equations

3. The attempt at a solution
$iz^2+2z-3i=0, z^2+(2/i)z-3=0$,using quadratic formula →$(-2/i± √8)/2 , z= √2+1/i$ and $√2-1/i$ is this correct?

Last edited by a moderator: May 26, 2016
2. May 26, 2016

### SteamKing

Staff Emeritus
I think in order to answer in the form x + i y, you need to rationalize your answers above.

https://mathway.com/examples/Linear.../Rationalizing-with-Complex-Conjugates?id=272

Last edited by a moderator: May 26, 2016
3. May 26, 2016

### chwala

ok we have $√2+1/i= (√2i+1)/i$ multiplying numerator and denominator by i we get $(-√2+1i)/-1 = √2-i$and → $√2+i$ respectively

4. May 26, 2016

### SteamKing

Staff Emeritus
Double check your arithmetic for the z-values which you calculated.

Last edited by a moderator: May 26, 2016
5. May 26, 2016

### chwala

still getting $(-√2+i)/-1= √2-i$

6. May 26, 2016

### ehild

Up to z=(-2/i± √8)/2 it is correct. Dividing by 2, you get -1/i ±√2. What is -1/i? Remember i was defined so i2=-1.

7. May 26, 2016

### Ray Vickson

If I were doing it I would first multiply the original equation by $-i$ to get $z^2 -2 i z -3=0$, which is a tinier bit simpler to work with.

8. May 26, 2016

### Math_QED

In fact you can check your answers by plugging in the solutions you found for z. Another way is to let wolframalpha solve it for you.

9. May 27, 2016

### chwala

Ray your approach to me looks good.
$z^2-2iz-3=0, ⇒((-2i±√(-4+12))/2⇒(-2i±2√2)/2⇒√2±i$
$z=√2+i, √2-i$
which is still the same as the solution i got earlier.

10. May 27, 2016

### SteamKing

Staff Emeritus
Ooooh! It's so subtle, you're still missing it!

You got $z = \frac{(-2i\, ±\, 2\sqrt{2})}{2}$ which you say is the same as $z = \sqrt{2}\, ±\, i$

11. May 27, 2016

### chwala

lol$z=-i+√2, z=-i-√2$ ⇒$z=√2-i, z=-√2-i$
Mathway gives solution as $z=√2-i$

12. May 27, 2016

### chwala

on another point, if i hadn't multiplied the original equation by $-i$, are we saying that we can't get alternative solution(s)? just a thought, like the way i had done it, or are we going to get the same solution as found in post11?

13. May 27, 2016

### chwala

post 9 looks wrong to me ......using the quadratic formula:, $z=(2i±2√2)/2$
and subsequently everything following from it.

14. May 27, 2016

### chwala

from post 13:, $z=i±√2$ which agrees with my earlier post number 3. Advice please?

15. May 27, 2016

### SteamKing

Staff Emeritus
$z = i\, ±\, \sqrt{2}$ (Post 14) is not the same as $z=\sqrt{2}\,±\,i$ (Post 3).

You can't flip where the ± goes in the expression.

16. May 27, 2016

### chwala

ok we have:, either $z=√2+i, z=-√2+i$

17. May 27, 2016

### chwala

look at my post 1;, it should be correct, was my rationalization wrong? this is how i did it,
the original equation, $iz^2+2z-3i=0,$ i divided each term by $i$, i got ,$z^2+(2/i)z-3=0$, on using the quadratic formula i got $z=((-2/i)±√(-4-4(1)(-3))/2, →((-2/i)±2√2)/2→(-1/i)±√2$⇒
$z=(-1/i)+√2, z=(-1/i)-√2$,
⇒$z=(-1+√2 i)/i, z=(-1-√2 i)/i$,
on multiplying the numerator and denominator by $i$, we have
$z=-i-√2, z=-i+√2$⇒ $z=-√2-i, z=√2-i$
mathway gives solution as $z=√2-i$, that is why i had mentioned earlier in post (12), if there is a possibility of many solutions to this problem dependant on the approach used.

Last edited: May 27, 2016
18. May 27, 2016

### SteamKing

Staff Emeritus
There's nothing wrong with rationalizing $\frac{1}{i}$ by multiplying that term by $\frac{i}{i}$.

That shouldn't change the fact that the quadratic formula gave $z = (\frac{\frac{-2}{i}±\sqrt{8}}{2})$.
You can rationalize the imaginary term without affecting the sign of the real term.

19. May 27, 2016

### chwala

steam thanks a lot, kindly look at my post number 1 and post number 3. Which means i was correct from that point.

20. May 27, 2016

### SteamKing

Staff Emeritus
That's just it. The problem starts in Post #1 and carries over into Post #3.