solving roots of complex numbers

[Relz]

(2 + i)z² - (l+i)z + 2i = 0

I have multiplied the z through each bracket but then i'm stuck. any ideas?

 

[burakaltr]

(2 + i)z² - (l+i)z + 2i = 0

I have multiplied the z through each bracket but then i'm stuck. any ideas?

Take z = a + b i

then you'll end up smt. like

f(a,b) + g(a,b)*i = 0 + 0*i

Take f(a,b)=g(a,b)=0

 

[Relz]

Take z = a + b i

then you'll end up smt. like

f(a,b) + g(a,b)*i = 0 + 0*i

Take f(a,b)=g(a,b)=0

I'm not quite sure how to get there. Where did f and g come from and how do I isolate z?

 

[burakaltr]

(2 + i)z² - (l+i)z + 2i = 0

I have multiplied the z through each bracket but then i'm stuck. any ideas?

If z is a complex number , it ought to be smt. like a+bi=z

z**2=a**2+2abi-b**2 Right ?

then


( 2 + i ) ( a**2+2abi-b**2)-(1+i)(a+bi)+2i=0


2a**2+4abi-2b**2+a**2 i -2ab -i b**2 -[(a+bi+ai-b)+2i=0


take reals and i-containing terms side by side

2a**2-2b**2-2ab-a+b=0 Right ?

AND

4ab+a**2-b**2-b-a+2=0

 

[Relz]

If z is a complex number , it ought to be smt. like a+bi=z

z**2=a**2+2abi-b**2 Right ?

then


( 2 + i ) ( a**2+2abi-b**2)-(1+i)(a+bi)+2i=0


2a**2+4abi-2b**2+a**2 i -2ab -i b**2 -[(a+bi+ai-b)+2i=0


take reals and i-containing terms side by side

2a**2-2b**2-2ab-a+b=0 Right ?

AND

4ab+a**2-b**2-b-a+2=0

I am supposed to solve for z

 

[burakaltr]

I am supposed to solve for z

Well, You are supposed to find a and b which in return constitute in [z = a + b i] fashion

PS: I hope to come out with a better sol'n towards the end of the day . I may be posting it soon

Best of Luck !

 

[soroban]

Hello, Relz!

\(\displaystyle (2 + i)z^2 - (1+i)z + 2i \:=\: 0\)

You have a quadratic equation.
How about using the Quadratic Formula?

\(\displaystyle z \;=\;\dfrac{(1+i) \pm \sqrt{(1+i)^2 - 4(2+i)(2i)}}{2(2+i)} \;=\;\dfrac{(1+i) \pm\sqrt{8-14i}}{2(2+i)} \)

 

[galactus]

Alas, that's exactly what I said, Soroban, but no one paid any attention.

 

[burakaltr]

Could Someone Please Provide the Proof that the Quadratic Formulae would work for an equation having complex numbers as coefficients .

Thank You !

 

[Deleted member 4993]

Could Someone Please Provide the Proof that the Quadratic Formulae would work for an equation having complex numbers as coefficients .

Thank You !

There is no reason not to work. Derivation of quadratic formula requires completing the square and finding square root - which is valid for all numbers (real or complex).

 

[burakaltr]

There is reason not to work. Derivation of quadratic formula requires completing the square and finding square root - which is valid for all numbers (real or complex).

Yes, Thank You, but Still how can one prove this in mathematical premises?

 

[Deleted member 4993]

Yes, Thank You, but Still how can one prove this in mathematical premises?

I suggest, you go through the derivation - and point out which step will not be valid for complex numbers. It is valid for sin(Θ), cos(Θ), ln(x), ex - why not for complex coefficients.

(a + b)² = a² + 2ab + b²

is valid for all a & b - real and complex. You can prove that by using FOIL.

Thus I do not see any step in derivation that will exclude complex coefficients.

 

[daon2]

I suggest, you go through the derivation - and point out which step will not be valid for complex numbers. It is valid for sin(Θ), cos(Θ), ln(x), ex - why not for complex coefficients.

(a + b)² = a² + 2ab + b²

is valid for all a & b - real and complex. You can prove that by using FOIL.

Thus I do not see any step in derivation that will exclude complex coefficients.

Exactly, nothing in the standard proof uses a property of the real numbers which does not hold for complex numbers (there are some properties for elements in R which do not hold in C).

Here is the proof I mean, assume a is not 0 then:

\(\displaystyle ax^2+bx+c \,\,=\,\, 0 \iff x^2 + \frac{b}{a}x \,\,=\,\, -\frac{c}{a}\,\, \iff \)

\(\displaystyle x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} \,\,=\,\, \frac{-4ac + b^2}{4a^2} \,\,\iff \)

\(\displaystyle (x+\frac{b}{2a})^2 \,\,= \,\,\frac{-4ac + b^2}{4a^2} \,\,\iff\)

\(\displaystyle x+\frac{b}{2a} \,\,= \,\,\pm \frac{\sqrt{-4ac + b^2}}{2a}\)

Then finish from there.

edit: I'm not so sure that simplifying the answer in the complex case is an easy task. It might be that the other method could be easier.

 

[Deleted member 4993]

edit: I'm not so sure that simplifying the answer in the complex case is an easy task. It might be that the other method could be easier.

using a + ib = r*e^iΘ

it could be done fairly easily with couple of tan^-1 and square roots.