View Full Version : Quadratic roots

Bobby Jones

07-08-2011, 05:33 AM

Morning people, Ive been asked to find the quadratic roots of 65.53-j47.61, I need to express the answer in rectangular form to 3 decimal places.

First I turned 65.53-j47.61 into polar form which makes: 81<324 degrees.

From here do I simply add and takeaway 90 degrees from 324 so I'd get

81<234 degrees

81<144 degrees

81<54 degrees

I'm very stuck as this dosent seem correct.

tkhunny

07-08-2011, 07:19 AM

For starters, how did you get three?

Second, the magnitude should not be the same. 9<{something}

Give it another go.

mmm4444bot

07-08-2011, 11:25 AM

find the quadratic roots of 65.53 - j47.61

I don't understand what this means. :(

This Complex number and its conjugate are the roots of the quadratic polynomial:

x^2 - 131.060x + 6560.893

I cannot find any information about quadratic roots of Complex numbers.

I turned 65.53 - j47.61 into polar form which makes: 81<324 degrees.

This polar form looks correct to me, but again I'm not sure what you're trying to do.

Bobby Jones

07-08-2011, 03:12 PM

It asks me to find the quadratic roots of 65.53-j47.61. And to express it in rectangular form.

Where should I start? Should I use the quadratic formula and work backwards?? Dont solve it for me, just point me in the right direction please.

Thankyou

tkhunny

07-08-2011, 06:39 PM

You could do that, but DeMoivre's formula is much quicker.

Try it the hard way:

A "Square Root" of 65.53 - 47.61i is a+bi, where a and b are Real.

(a+bi)^2 = (a^2 - b^2) + i(2ab).

Thus:

a^2 - b^2 = 65.53 and

2*a*b = 47.61

Solve the non-linear system.

You should be able to check a^{2} + b^{2} = \sqrt{65.53^{2}+47.61^{2}}

Bobby Jones

07-09-2011, 05:39 AM

I've been going though the De Moivres formula and I've come up with these answer:

The quadratic roots for 65.53-j47.61 are,

2.78+j8.56

and

-2.78-j8.56

Please see if I'm correct as I could do with an experts opinion.

Thanks guys

tkhunny

07-09-2011, 07:21 AM

You can check. Use that formula on the end. That will be a magnitude check.

Plot them or calculate the angles between various vectors. It should make sense.

mmm4444bot

07-09-2011, 02:04 PM

The quadratic roots for 65.53-j47.61 are,

2.78 + j8.56

-2.78 - j8.56

Squaring each of your roots must yield the original Complex number, yes?

You had better check that.

Also, you have not rounded to the correct number of decimal places.

I would still like to know why the requested "roots" of that Complex number are called "quadratic roots". :?

If, instead, they were to ask for the principal square root of the given Complex number, then that would make sense to me.

By the way, you started out converting to polar form, where r = 81.

Here's a formula for the principal square root of a Complex number using r:

\sqrt{x+iy} = \sqrt{\frac{r + x}{2}} \pm i \sqrt{\frac{r - x}{2}}

where the sign of the imaginary part of the root is chosen to be the same as the sign of the imaginary part of the original number.

Cheers ~ Mark 8-)

Bobby Jones

07-11-2011, 12:30 PM

Mark,

I've tried the formulas you gave and with r = 81, x= 65.53 and y = -47.61 I get the answer 8.559 +- i6.945

But using the De Moviers formula I am now getting the answer to the two roots as +- 7.281 +- i5.290

Can someone work it through please and tell me which one is the correct answer.

mmm4444bot

07-11-2011, 03:00 PM

Mark, I've tried the formulas you gave

Huh? I gave only one formula.

I get the answer 8.559 +- i6.945

If you are trying to find the principal square root of the given Complex number, then there is only ONE answer.

(I don't yet know for sure what they want you to do. The formula that I posted is based entirely on a GUESS.)

If you believe that they want you to report the principal square root, then you may not write "+-". There is only one principal square root. You must choose the sign according to the sign of the imaginary part of the original Complex number.

There are different ways to calculate the principal square root; I posted one way, and the reason why that post is based on a guess is because the phrase "quadratic roots of a Complex number" has no meaning (to me).

The number 6.945 is not correct. Check your calculations on the following.

\sqrt{\frac{81 - 65.53}{2}}

Your result for the Real part is correct (8.559).

But using [de Moivre's] formula I am now getting the answer to the two roots as +- 7.281 +- i5.290

What you have typed here represents four numbers, not two. :?

I cannot comment on your work with this formula.

I still do not even know what your exercise wants.

Bobby Jones

07-12-2011, 05:48 AM

Mark,

I now get 8.559 + i2.78

But in your formula where does the y part go when you expand out?

mmm4444bot

07-12-2011, 02:26 PM

Mark,

I now get 8.559 + i2.78

The number in red is not rounded to the correct number of decimal places. You're still being sloppy.

But in your formula where does the y part go when you expand out?

The phrase in red is nonsensical.

There is no y in the expression for the principal square root of x + y*i that I posted; hence, you do not need to do anything with y.

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