Complex numbers : quadratic equation

[chwala]

Homework Statement

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,

Homework Equations

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?

 

[SteamKing]

Homework Statement

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,

Homework Equations

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?

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

 

[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

 

[SteamKing]

Homework Statement

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,

Homework Equations

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?

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

 

[chwala]

still getting $(-√2+i)/-1= √2-i$

 

[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.

 

[Ray Vickson]

Homework Statement

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,

Homework Equations

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?

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.

 

[member 587159]

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.

 

[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.

 

[SteamKing]

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.

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 $z = \sqrt{2}\, ±\, i$

 

[chwala]

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

 

[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?

 

[chwala]

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

 

[chwala]

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

 

[SteamKing]

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

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

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

 

[chwala]

ok we have:, either $z=√2+i, z=-√2+i$

 

[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.

 

[SteamKing]

look at my post 1;, it should be correct, was my rationalization wrong? this is how i did it,

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.

 

[chwala]

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

 

[SteamKing]

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

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

You had:

... using quadratic formula →(-2/i± √8)/2 ,

which is correct. Then you wrote

$z= √2+1/i$ and $√2-1/i$ is this correct?

which is not correct, and not because of how you rationalized the imaginary term in z above.

It's not correct because you moved the ± from in front of the square root and put it in front of the imaginary term.

It's like saying 4 ± 6 = 6 ± 4. You can check this example and see that while 4 + 6 = 6 + 4, the converse 4 - 6 ≠ 6 - 4. It's a very subtle distinction, but one which nevertheless must be made.

 

[chwala]

agreed i can see the mistake.

 

[chwala]

I can see my mistake, thanks again.

 

[SteamKing]

I can see my mistake, thanks again.

You're welcome. Good luck.