complex numbers question! (1 Viewer)

[felixcthecat]

hii.. well umm i was reviewing my complex numbers and i've found that i've completely forgotten the method to solve the following type of question.. can someone do that example so i'd kno? thanx!

z^2 - (1-i)z + 7i-4 = 0

i looked in my notes but the method seems rather stupid cuz it went into z^4.. none-the-less, can someone do it plz? thanx!

 

[Riviet]

Three words: "Use Quadratic Formula"

To find square roots of complex numbers e.g sqrt(1+i), let (a+ib)²=1+i

expand LHS, equate real and imaginary parts from both sides to form two equations in a and b. Simultaneously solve them and substitute the values of a and b into a+ib to get your 2 complex roots.

 

[buchanan]

 

[felixcthecat]

thanx buchanan!!.. do u hav worked solutions for the whole coroneos book or sumthing? =D

haha riviet.. i got to the step where u told me to equate sqrt of the complex number and stuff... but then since i was looking at my notes, i didn't think and tried to find the x and y of (x + iy) haha.. my teacher makes things so difficult when they're so easy...

thanks!~

edit: buchanan.. how'd u rememeber it was coroneos 1C question 9 part 3?!? ...

 

[Rax]

Ok I don't think I have done any complex number questions like that, so can someone please explain the formula bit with the "square root of w" section.
I get the quadratic eqtn bit obviously lol
Thanks

 

[buchanan]

In my solution, w=16-30i.

 

[pLuvia]

Ok well to find the square root of 16-30i you let z² = 16-30i (z=x+iy)

z²=16-30i
(x+iy)²=16-30i
Equate Re and Im parts

x²-y²=16 -------------- 1
2xy=-30
y=-15/x ---------------2

Using simulataneous equations solve for x and y

you should end up having
x=5
y=-3

Which you sub into the general equation of a complex number

therefore the square root of 16-30i = 5-3i

Hope that helped

 

[buchanan]

It's easier if you use the formula:

 

[Riviet]

Buchanan- what does the isgn bit mean in the formula?

 

[pLuvia]

It's easier if you use the formula:

Never learnt the formula Only learnt it by equating Re and Im parts and DeMoivre's theorem

 

[Riviet]

The formula is valid, just not in the syllabus. However, use at your own decision and risk.

 

[buchanan]

It is in the syllabus.

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"

And if you do as the syllabus says and actually solve those equations, you get

sgn(b) is the sign of b. So if b > 0, sgn(b)=+1, if b < 0, sgn(b)=-1. If b=0, sgn(b)=0, but then there's no need for the formula.

Using this formula is a much faster method and so as far as risk is concerned, there's much less chance of making a mistake if you use the formula.

If you use the formula, you can get the answer in 1 line.

Other examples:

When you solve quadratics, do you use completing the square every time? That would be stupid! There's a formula which makes the solution much faster, &there4; use it!

It's the same for finding square roots of complex numbers. There's a formula which makes the solution much faster, &there4; use it!

(and dare I say, it's the same for <a href="http://community.boredofstudies.org/238/appreciating-beauty-elegance-extracurricular-topics/100243/third-derivative-method.html">third derivative method</a>!)

 

[felixcthecat]

now that i look back at your first answer buchanan.. y'd u need the formula? .. i thought u could do it without the formula exactly the same way u did it?.. or am i just subconciously using that formula ^0^

 

[buchanan]

My solution to the Coroneos question used 2 formulae simultaneously. Square roots of a complex number, and quadratic formula.

You don't have to use the quadratic formula to solve a quadratic, but it's easier in your example if you do.

Likewise, you don't have to use the formula for the square roots of a complex number, but again, in your example, it makes it easier if you do.

 

[Rax]

ok Sweet.. I just wasn't too sure as I hadnt seen that formula before. Does anyone know its derivation (Guessing thats out of syllabus for sure) because I would like to find how they got it.
Thanks guys

 

[buchanan]

It's not hard to derive.

Square both sides:

Or alternatively just follow what pLuvia and Riviet said, but in the general case (a bit longer, but it still works).

 

[felixcthecat]

Ok well to find the square root of 16-30i you let z² = 16-30i (z=x+iy)

z²=16-30i
(x+iy)²=16-30i
Equate Re and Im parts

x²-y²=16 -------------- 1
2xy=-30
y=-15/x ---------------2

Using simulataneous equations solve for x and y

you should end up having
x=5
y=-3

Which you sub into the general equation of a complex number

therefore the square root of 16-30i = 5-3i

Hope that helped

i'm also wondering... if u don't use buchanan's equation.. how would u derive x=5 and y=-3 from ..
x²-y²=16
and
2xy=-30

cuz u'd come out with.. x² - 15²/x² = 16

which would come out with x⁴ - 16x² - 15 = 0

how'd u go from there?.. i would hav thought u didn't need to go there and i've missed something realli vital... did i?

 

[buchanan]

which would come out with x⁴ - 16x² - 15 = 0

You missed the square (which I made red):
x⁴-16x²-15²=0.

 

[felixcthecat]

ohh haha.. stupid me =.=.. the legacy of being lazy all holidays ^^''

 

[pLuvia]

Silly mistakes, just watch out for them