How to solve w^2=-3-4i

[mathsenjoy]

Ok so i have gotten as far to get the correct answers and used the quadratic equation. What i dont understand is how the final answer is w=1-2i or w=-1+2i
why is it in the answers one number becomes postive and the other negative???
like in w=1-2i why is 1 positive and why is 2 negative
I understand that they can both be negative or postive when using the quadratic equation but i didnt understand how one number can be positive and the other negative.
Please help
If i put this in the wrong thread please tell me, i am just learning maths properly at an older age

 

[MarkFL]

I would write:

\(\displaystyle w^2=-3-4i=(1-4)-4i=1-4i+4i^2=(1-2i)^2\)

Thus:

\(\displaystyle w=\pm(1-2i)\)

 

[Dr.Peterson]

Ok so i have gotten as far to get the correct answers and used the quadratic equation. What i dont understand is how the final answer is w=1-2i or w=-1+2i
why is it in the answers one number becomes postive and the other negative???
like in w=1-2i why is 1 positive and why is 2 negative
I understand that they can both be negative or postive when using the quadratic equation but i didnt understand how one number can be positive and the other negative.
Please help
If i put this in the wrong thread please tell me, i am just learning maths properly at an older age

I'd like to see your work in order to tell you how to do your final step. There are many ways to solve a problem, which may reveal details in different ways.

When you say you "used the quadratic equation", do you mean the quadratic formula? Please show details.

 

[pka]

Ok so i have gotten as far to get the correct answers and used the quadratic equation. What i dont understand is how the final answer is w=1-2i or w=-1+2i
why is it in the answers one number becomes postive and the other negative???
like in w=1-2i why is 1 positive and why is 2 negative
I understand that they can both be negative or postive when using the quadratic equation but i didnt understand how one number can be positive and the other negative.

Hello Mathsenjoy. Your statement about positive or negative complex numbers tells me that you may be not be ready to understand the answer. Now that is not any knock on you but rather a knock on mathematics education. There are no positive or negative complex numbers.
The complex numbers do not form an ordered field. Therefore it is not possible to have numbers compared to zero.
If [imath]n\in\mathbb{Z}^+~\&~z\in\mathbb{C}[/imath] there are [imath]n[/imath] [imath]n^{th}\text{roots of }z[/imath].
Thus to your question: there are two square-roots of the complex number [imath]1+2\bf i[/imath].
To find the roots we must write [imath]1+2\bf i[/imath] in polar form: [imath]1+2{\bf i}=\sqrt{5}e^{{\bf i}\theta}[/imath] where [imath]\theta=\arctan(2)[/imath].
From there we use that to find roots.


[imath][/imath][imath][/imath][imath][/imath]

 

[mmm4444bot]

Your statement about positive or negative complex numbers tells me that you may be not be ready to understand …

There are no positive or negative complex numbers.

I don't read the op like that. (I think you've misinterpreted a pronoun.) In their example, mathsenjoy specifically asked about a and b in a+bi.

why is 1 positive and why is 2 negative

[imath]\;[/imath]

 

[Otis]

w=1-2i
why is 1 positive and why is 2 negative[?] I understand that they can both be negative or positive

Hi mathsenjoy. Complex numbers take this form: [imath]\color{blue}a+b \cdot i[/imath], where symbols [imath]a[/imath] and [imath]b[/imath] each represent a Real constant (so each may be positive or negative or zero). Symbol [imath]i[/imath] is [imath]\sqrt{\text{-}1}[/imath].

The signs of constants [imath]a[/imath] and [imath]b[/imath] do not need to be the same. For Complex number [imath]w[/imath] in this exercise, there is no special reason why the signs of [imath]a[/imath] and [imath]b[/imath] are different. Those are one pair of values that yield [imath]\text{-}3\,–\,4i[/imath] when [imath]w[/imath] is squared. If the signs were the same, then [imath]w^2[/imath] would be [imath]\text{-}3+4i[/imath], instead. (In other words, a different exercise.)

As Mark showed, the two square roots are 1–2i and -(1–2i). We call 1–2i the principle square root of w², and the secondary square root is -(1–2i).

I, too, am curious to see which method you'd used, to find the values of [imath]a[/imath] and [imath]b[/imath]. Please share.
[imath]\;[/imath]

 

[mathsenjoy]

Hi, thanks for all response, I hope it's okay that i respond here to all messages. Before i start I just want to state that I grew up as a child without formal education for sometime so I have been learning a lot of basics.



calculation I used (based on my UK A level textbook guidelines):

w^2=-3-4i

Let w=a+bi

(a+bi)^2= -3-4i

a^2+2abi+a^2b^2=-3-4i

(a^2-b^2)+2abi=-3-4i

Real numbers: a^2-b^2=-3

Imaginary number: 2ab=-4
ab=-2
-ab=2
-b=2/a

Substitute imaginary into real:
a^2-(2/a)^2=-3
a^2-4/a^2=-3

a^4-4=3a^2
a^4+3a^2-4=0

Use quadratic formula:
a^2=-3+-square root of 3^2-4(1)(-4)/2(1)=+-1

a=+-1

-b=2/1

w=1-2i or w=-1+2i

SORRY EVERYONE I JUST REALISED I UNDERSTAND NOW WHEN I WENT THROUGH IT JUST NOW
hope i didn't waste everyone's time, when I went through it again I understand why b is minus and a is plus and vice versa

Thank you everyone for your help, you have all been so kind to me, it is very heartwarming to see such a great helpful bunch of people!

 

[Otis]

hope i didn't waste everyone's time

Hi there! This thread is not a waste at all. It helped you to gain a better understanding, and it serves future readers.

Thanks for posting your work — your method is what I would've done, except for the quadratic formula. We're happy you understand the answer better. (That's what happens, the more we practice.)

I have some comments about your work, and I suspect most of them concern only typing goofs. I'll also include (in red) some examples of common ways we show math when using a keyboard.

a^2+2abi+a^2b^2=-3-4i

The factor i^2 or -1 is missing, on the left-hand side. Either of the following would do.

a^2 + 2abi + a^2*b^2*i^2 = -3–4i

a^2 + 2abi – a^2*b^2 = -3–4i

Real numbers: a^2-b^2=-3

Imaginary number: 2ab=-4

This comment has to do with terminology. There's only one Real part in [imath]w[/imath]. Also, all Imaginary numbers contain i. In the expression a+bi, we have noun phrases for the parameters [imath]a[/imath] and [imath]b[/imath]. We call [imath]a[/imath] the 'Real part' and [imath]b[/imath] the 'Imaginary part'. (Both "parts" are Real numbers.) Therefore, expressions 2abi and -4i each represent an Imaginary number, while your expressions 2ab and -4 above represent the Imaginary part, of w.

Real part of w: a^2 – b^2 = -3

Imaginary part of w: 2ab = -4

-b=2/a

We're trying to find b, so it's handy to get an expression for b instead of -b.

b = -2/a

a^4-4=3a^2

Missing sign.

a^4 – 4 = -3a^2

a^2=-3+-square root of 3^2-4(1)(-4)/2(1)=+-1

a = +-1

We know [imath]a[/imath] is a Real number, so its square cannot equal -1. We ignore that negative result, by stating a^2=1.

a^2 = -3 +/- sqrt[3^2 – 4(1)(-4)]/[2(1)] = +/-1
or
a^2 = -3 ± √[3^2 – 4(1)(-4)]/[2(1)] = ±1

Then
a^2 = 1
a = +/-1

---

Here are alternative ways to find [imath]a[/imath] and [imath]b[/imath].

It's easy to factor the quadratic polynomial x^2+3x–4 as (x+4)(x–1) because 4 and -1 are a pair of numbers whose sum is 3 and whose product is -4. Thus, we can factor our quadratic-form polynomial below the same way.

a^4 + 3a^2 – 4 = 0

(a^2 + 4)(a^2 – 1) = 0

The zero-product property tells us:

a^2 = -4 [imath]\quad \text{OR} \quad [/imath] a^2 = 1

Ignore the negative square, so a = ±1.

We could also have taken the following approach earlier (because 3 is a prime number):

a^2 – b^2 = -3

Factor the difference of squares.

(a + b)(a – b) = -3

Let (a + b) = 3 and (a – b) = -1. Then add those equations, to eliminate b.

a + b = 3
a – b = -1
-----------
2a + 0 = 2

a = 1

(In your exercise, it doesn't matter which way you assign values to a+b and a–b to obtain the product -3. The end result will be the same. And for any method, when finding b=-2/a, it doesn't matter if you use a=1 or a=-1.)


[imath]\;[/imath]

 

[farhan]

To solve the equation w^2 = -3 - 4i, we can use the following steps:

1: Rewrite -3 - 4i in polar form We can find the magnitude r and argument θ of -3 - 4i using the formulas: r = sqrt((-3)^2 + (-4)^2) = 5 θ = arctan(-4/-3) = -53.13° (or -0.93 radians) So, -3 - 4i can be written as 5cis(-53.13°) or 5cis(-0.93).

2: Find the square root of -3 - 4i We can find the square roots of 5cis(-0.93) by using the formula: w = ± sqrt(r) * cis(θ/2) where the ± sign indicates that there are two possible solutions.

Substituting the values for r and θ, we get: w = ± sqrt(5) * cis(-0.93/2) w = ± sqrt(5) * cis(-0.465)

Using Euler's formula, cis(x) = cos(x) + isin(x), we can rewrite w as: w = ± sqrt(5) * (cos(-0.465) + isin(-0.465)) w = ± sqrt(5) * (0.891 - 0.454i)

Therefore, the solutions to w^2 = -3 - 4i are: w = ± sqrt(5) * (0.891 - 0.454i)

 

[Dr.Peterson]

To solve the equation w^2 = -3 - 4i, we can use the following steps:

1: Rewrite -3 - 4i in polar form We can find the magnitude r and argument θ of -3 - 4i using the formulas: r = sqrt((-3)^2 + (-4)^2) = 5 θ = arctan(-4/-3) = -53.13° (or -0.93 radians) So, -3 - 4i can be written as 5cis(-53.13°) or 5cis(-0.93).

2: Find the square root of -3 - 4i We can find the square roots of 5cis(-0.93) by using the formula: w = ± sqrt(r) * cis(θ/2) where the ± sign indicates that there are two possible solutions.

Substituting the values for r and θ, we get: w = ± sqrt(5) * cis(-0.93/2) w = ± sqrt(5) * cis(-0.465)

Using Euler's formula, cis(x) = cos(x) + isin(x), we can rewrite w as: w = ± sqrt(5) * (cos(-0.465) + isin(-0.465)) w = ± sqrt(5) * (0.891 - 0.454i)

Therefore, the solutions to w^2 = -3 - 4i are: w = ± sqrt(5) * (0.891 - 0.454i)

No, your solutions are approximations to the exact solutions [imath]\pm(2-i)[/imath], and they are also wrong! If you square them, you get [imath]3-4i[/imath], not [imath]-3-4i[/imath].

This happened because your angle -53.13° is in the wrong quadrant.

 

[mathsenjoy]

Hi there! This thread is not a waste at all. It helped you to gain a better understanding, and it serves future readers.

Thanks for posting your work — your method is what I would've done, except for the quadratic formula. We're happy you understand the answer better. (That's what happens, the more we practice.)

I have some comments about your work, and I suspect most of them concern only typing goofs. I'll also include (in red) some examples of common ways we show math when using a keyboard.


The factor i^2 or -1 is missing, on the left-hand side. Either of the following would do.

a^2 + 2abi + a^2*b^2*i^2 = -3–4i

a^2 + 2abi – a^2*b^2 = -3–4i


This comment has to do with terminology. There's only one Real part in [imath]w[/imath]. Also, all Imaginary numbers contain i. In the expression a+bi, we have noun phrases for the parameters [imath]a[/imath] and [imath]b[/imath]. We call [imath]a[/imath] the 'Real part' and [imath]b[/imath] the 'Imaginary part'. (Both "parts" are Real numbers.) Therefore, expressions 2abi and -4i each represent an Imaginary number, while your expressions 2ab and -4 above represent the Imaginary part, of w.

Real part of w: a^2 – b^2 = -3

Imaginary part of w: 2ab = -4


We're trying to find b, so it's handy to get an expression for b instead of -b.

b = -2/a


Missing sign.

a^4 – 4 = -3a^2


We know [imath]a[/imath] is a Real number, so its square cannot equal -1. We ignore that negative result, by stating a^2=1.

a^2 = -3 +/- sqrt[3^2 – 4(1)(-4)]/[2(1)] = +/-1
or
a^2 = -3 ± √[3^2 – 4(1)(-4)]/[2(1)] = ±1

Then
a^2 = 1
a = +/-1

---

Here are alternative ways to find [imath]a[/imath] and [imath]b[/imath].

It's easy to factor the quadratic polynomial x^2+3x–4 as (x+4)(x–1) because 4 and -1 are a pair of numbers whose sum is 3 and whose product is -4. Thus, we can factor our quadratic-form polynomial below the same way.

a^4 + 3a^2 – 4 = 0

(a^2 + 4)(a^2 – 1) = 0

The zero-product property tells us:

a^2 = -4 [imath]\quad \text{OR} \quad [/imath] a^2 = 1

Ignore the negative square, so a = ±1.

We could also have taken the following approach earlier (because 3 is a prime number):

a^2 – b^2 = -3

Factor the difference of squares.

(a + b)(a – b) = -3

Let (a + b) = 3 and (a – b) = -1. Then add those equations, to eliminate b.

a + b = 3
a – b = -1
-----------
2a + 0 = 2

a = 1

(In your exercise, it doesn't matter which way you assign values to a+b and a–b to obtain the product -3. The end result will be the same. And for any method, when finding b=-2/a, it doesn't matter if you use a=1 or a=-1.)


[imath]\;[/imath]

thank you and yes some were typing errors but others were just me not fully understanding the method, i will attempt some more questions soon. I'm very happy for all the responses!

 

[Steven G]

-b=2/a

Substitute imaginary into real:
a^2-(2/a)^2=-3

You clearly replaced b with -2/a. The problem is that does not equal -2/a !!