Finding square roots of complex numbers (1 Viewer)

[YBK]

Hey,

I can find square roots of complex numbers using the 'normal' way... (equating the real and imaginary parts) but now that I am doing work from the Terry Lee book, there's a different method called completing the square.

I can do a few examples using that method, and for those ones I get the answer much faster than normal, but I don't really understand the explanation of it in the book...

for example, how would you find the roots of - 3 + 4i using complete the square method ?


and btw, is that method preferable to the other? Is it acceptable for it to be used in an exam?



thanks!!!

 

[pritnep]

Await Riviet's intelligent mathematical advice… ( sorry I can’t help, don’t really have a mathematical brain - but I do for computers )

 

[YBK]

Await Riviet's intelligent mathematical advice… ( sorry I can’t help, don’t really have a mathematical brain - but I do for computers )

True, Riviet = greatness

 

[Riviet]

I have never used that method before to be honest

If you're using terry lee's 6th edition, then turn to page 31 and there's an example fully explained in method 2.

It't not really completing the sqrare but more like breaking up the real part and imaginary part further. It has a very familiar procedure to factorising quadratics. I still like just simply solving the two equation simultaneously.

If i figure out how it works, i'll post up a worked example in my own words.

 

[YBK]

Thanks Riviet! I have the 5th edition, and its explanation is only around 4 lines..

 

[insert-username]

Either method should be equally acceptable in the exam. Both get you the correct answer, after all.

EDIT: See you teacher if you need more help, but it's usually better to stick with the method you know better.


I_F

 

[YBK]

Either method should be equally acceptable in the exam. Both get you the correct answer, after all.

EDIT: See you teacher if you need more help, but it's usually better to stick with the method you know better.


I_F

yeah, I plan to ask her when we get back - but she never taught us the short way....

 

[Riviet]

Okay, i understand it quite well now. It makes alot of sense.

I'll do your example, -3+4i :

I don't know about you but you should be able to write down the two equations straight off, using -3+4i = a²-b²+2abi and equating the Re and Im parts to obtain:

a²-b²=-3
2ab=4

Now the first thing we do is check that the b in the complex number is even because this will correspond to the 2ab when we take out the two from the even number. So in this example b=4, so we write 2x2, which corresponds to 2ab and then if you cancel the two from both you get ab=2, so we find two numbers that multiply to give 2 (a and b), as well as the difference between their squares equals -3. Now we mentally work out that 1²-2²=-3 and the order here is important because if it was the other around we would get 3. equating the numbers in 1²-2²=-3 and a²-b²=-3, it should be obvious that a=1, and b=2. Therefore your answer is a+bi => 1+2i.
But it is a square root so we need to take the positive and negative of it, so the final answer is +(1+2i).
This can actually be all done in your head, so some working might allow us to use this in an exam or assessment. You could write:

sqrt(-3+4i) = sqrt(1²-2²+2x1x2xi)

We write the RHS like that because from:

(a+ib)²=a²-b²+2abi

We take the square root of both sides to obtain:

a+ib= +sqrt(a²-b²+2abi)

Notice how the 1's and 2's correspond to the a's and b's in the RHS, ie a²-b²+2xaxbxi.

One more thing. To find square roots of purely imaginary numbers such as 24i, we write it with the real part too, ie 0+24i and so a²-b²=0 and 2ab=24 and use the above procedure to work out the square root.

If you have any more questions, feel free to ask

 

[YBK]

Okay, i understand it quite well now. It makes alot of sense.

I'll do your example, -3+4i :

I don't know about you but you should be able to write down the two equations straight off, using -3+4i = a²-b²+2abi and equating the Re and Im parts to obtain:

a²-b²=-3
2ab=4

Now the first thing we do is check that the b in the complex number is even because this will correspond to the 2ab when we take out the two from the even number. So in this example b=4, so we write 2x2, which corresponds to 2ab and then if you cancel the two from both you get ab=2, so we find two numbers that multiply to give 2 (a and b), as well as the difference between their squares equals -3. Now we mentally work out that 1²-2²=-3 and the order here is important because if it was the other around we would get 3. equating the numbers in 1²-2²=-3 and a²-b²=-3, it should be obvious that a=1, and b=2. Therefore your answer is a+bi => 1+2i.
But it is a square root so we need to take the positive and negative of it, so the final answer is +(1+2i).
This can actually be all done in your head, so some working might allow us to use this in an exam or assessment. You could write:

sqrt(-3+4i) = sqrt(1²-2²+2x1x2xi)

We write the RHS like that because from:

(a+ib)²=a²-b²+2abi

We take the square root of both sides to obtain:

a+ib= +sqrt(a²-b²+2abi)

Notice how the 1's and 2's correspond to the a's and b's in the RHS, ie a²-b²+2xaxbxi.

One more thing. To find square roots of purely imaginary numbers such as 24i, we write it with the real part too, ie 0+24i and so a²-b²=0 and 2ab=24 and use the above procedure to work out the square root.

If you have any more questions, feel free to ask

yay!!!!!!

thank you so much!! I completely understand it now!!

 

[KeypadSDM]

As an aside, if posed with something nice to find the square root of [such as 1 + Sqrt[3]i], then you can use mod-arg square rooting.

E.g:

Sqrt[1 + Sqrt[3]i]
= Sqrt[2 . cis(2.n.pi + pi/3)]
= Sqrt[2] . cis(n.pi + pi/6)

Then taking n = 0 and n = 1 gives the 2 roots.

= Sqrt[2] . cis(pi/6) or Sqrt[2] . cis(7pi/6)
Then break it down.

Although this looks a tad longer, if you solve the other simeltaneous equations directly [not by observation] then this comes out ALOT quicker.

But remember, it only works analytically for good arguments, and numerically for all arguments [for a numerical answer, this method is always quicker].

 

[KeypadSDM]

The principal square root of x+iy is ETC.

Oh fine, just work out the formula and screw us all up. I don't think examiners will like you if you just plug in a formula.

 

[bananasmoothy]

I can find square roots of complex numbers using the 'normal' way... (equating the real and imaginary parts) but now that I am doing work from the Terry Lee book, there's a different method called completing the square.

It sounds as if you've never actually done basic equations via completing the square method before. Either that, or you're just applying it to the square root of complex numbers thing.
If you never done completing the square before, maybe you should go back and do some basic ones to fully understand it. Though I find it hard to believe you've never done it before...

 

[Riviet]

The principal square root of x+iy is

(with positive real part)

The other one is the negative of this.

VERY INTERESTING

-to look at-

I'll just forget about it now. Don't need these massive formulas to remember. Got enough anyway. :]

 

[YBK]

It sounds as if you've never actually done basic equations via completing the square method before. Either that, or you're just applying it to the square root of complex numbers thing.
If you never done completing the square before, maybe you should go back and do some basic ones to fully understand it. Though I find it hard to believe you've never done it before...

I've done it countless times before, although it is quite different for the complex numbers. haha, can't imagine maths without completing the square (for normal equations)

 

[YBK]

The principal square root of x+iy is

(with positive real part)

The other one is the negative of this.

lol, cool!!

although it seems to big and difficult for my liking.. i've just fell in love with my new learnt method! w00t w00t

 

[icycloud]

I agree that the formula looks a bit big, but most questions on square roots of complex numbers are contrived to give nice answers - in which case, the formula actually is very easy to use.

But you can't use it in exam situations anyway. 'Proper' working must be shown.

 

[YBK]

But you can't use it in exam situations anyway. 'Proper' working must be shown.

If proper working must be shown, then the completing the square method isn't too pretty....

 

[YBK]

See if you can prove the general formula.

the 1/2 coefficient makes sense because you have to make the imaginary part even...

 

[Riviet]

Nice observation YBK.

I'm gonna stick to the systematic way of doing it.

 

[A l]

Altough it is a quick method, some schools won't approve of it, because the formula is not mentioned in the syllabus and if you look closely at the applications, implications and considerations section of the complex number syllabus, it states:
"In finding the square roots of a + ib, the statement √(a + ib) = x + iy, where a, b, x, y are real, leads to the need to solve the equations x² - y² = a and 2xy = b"