Complex numbers : quadratic equation

In summary, when solving the equation iz^2+2z-3i=0, the quadratic formula is used to get two possible solutions for z, which are -i-√2 and -i+√2. These can also be written as √2-i and -√2-i, respectively. However, it is important to note that the ± sign in front of the square root should not be moved to the imaginary term.
  • #1
chwala
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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?
 
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  • #2
chwala said:

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
 
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  • #3
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
chwala said:

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.
 
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  • #5
still getting ## (-√2+i)/-1= √2-i##
 
  • #6
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.
 
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  • #7
chwala said:

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.
 
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  • #8
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.
 
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  • #9
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
chwala said:
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##
 
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  • #11
lol##z=-i+√2, z=-i-√2## ⇒##z=√2-i, z=-√2-i##
Mathway gives solution as ##z=√2-i##
 
  • #12
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
post 9 looks wrong to me ...using the quadratic formula:, ##z=(2i±2√2)/2##
and subsequently everything following from it.
 
  • #14
from post 13:, ##z=i±√2## which agrees with my earlier post number 3. Advice please?
 
  • #15
chwala said:
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.
 
  • #16
ok we have:, either ##z=√2+i, z=-√2+i##
 
  • #17
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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  • #18
chwala said:
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.
 
  • #19
steam thanks a lot, kindly look at my post number 1 and post number 3. Which means i was correct from that point.
 
  • #20
chwala said:
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:
chwala said:
... 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.
 
  • #21
agreed i can see the mistake.
 
  • #22
I can see my mistake, thanks again.
 
  • #23
chwala said:
I can see my mistake, thanks again.
You're welcome. Good luck.
 
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Related to Complex numbers : quadratic equation

1. What are complex numbers?

Complex numbers are numbers that consist of both real and imaginary parts. They are written in the form a + bi, where a is the real part and bi is the imaginary part, with i representing the square root of -1. Complex numbers are useful in solving certain types of equations, such as quadratic equations.

2. What is a quadratic equation?

A quadratic equation is a polynomial equation of the form ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable. It is called a quadratic equation because x is raised to the second power, making it a quadratic function. Quadratic equations have two solutions, or roots, which can be found using the quadratic formula.

3. How do complex numbers relate to quadratic equations?

Complex numbers can be used to find the roots of quadratic equations, even when the solutions are not real numbers. This is because the quadratic formula, which is used to solve quadratic equations, includes the square root of -1, which is represented by i. When solving for the roots of a quadratic equation, both real and complex solutions are considered.

4. Can a quadratic equation have only complex solutions?

Yes, a quadratic equation can have only complex solutions. This means that the solutions to the equation will be complex numbers and there will be no real solutions. This can happen when the discriminant, b^2 - 4ac, is negative. In this case, the solutions will be in the form of a + bi, where a and b are real numbers and i is the imaginary unit.

5. How are complex numbers used in real life?

Complex numbers are used in various fields of science, such as physics and engineering, to represent quantities that have both magnitude and direction. They are also used in signal processing, electrical engineering, and computer graphics. In mathematics, complex numbers are used to solve certain types of equations, such as quadratic equations, and they have important applications in calculus, number theory, and other branches of mathematics.

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