# Complex numbers : quadratic equation

Gold Member

## 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,

## 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?

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SteamKing
Staff Emeritus
Homework Helper

## 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,

## 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

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Gold Member
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
Staff Emeritus
Homework Helper

## 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,

## 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.

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Gold Member
still getting ## (-√2+i)/-1= √2-i##

ehild
Homework Helper
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.

chwala
Ray Vickson
Homework Helper
Dearly Missed

## 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,

## 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.

chwala and SammyS
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
Gold Member
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
Staff Emeritus
Homework Helper
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
Gold Member
lol##z=-i+√2, z=-i-√2## ⇒##z=√2-i, z=-√2-i##
Mathway gives solution as ##z=√2-i##

Gold Member
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?

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

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

SteamKing
Staff Emeritus
Homework Helper
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.

Gold Member
ok we have:, either ##z=√2+i, z=-√2+i##

Gold Member
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.

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SteamKing
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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.

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

SteamKing
Staff Emeritus
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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.

... 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.

Gold Member
agreed i can see the mistake.

Gold Member
I can see my mistake, thanks again.

SteamKing
Staff Emeritus