complex numbers (1 Viewer)

[tacogym27101990]

hey
im a little bit rusty at complex, can someone please give me a bit hint as to how to solve this?

thanks

 

[tommykins]

z = x + iy
x^2 +2ixy - y^2 + 2(x-iy) + 1 = 0
x^2 - y^2 + 2x -2iy + 2ixy + 1 = 0

equate imaginary and real to 0

 

[vds700]

hey
im a little bit rusty at complex, can someone please give me a bit hint as to how to solve this?

thanks

u tutoring 4 unit?

 

[TBK11]

thanks

wouldn't it be

z^2+2z x 1+1^2=0

(z+1)^2=0

{ (z+1)^2} ^1/2=0

2x1/2
(z+1) =0

(z+1)^2/2=0

(z+1)^1/1=0

(z+1)^1=0

z+1=0

(z+1)+(-1)=0+(-1)

(z+1)-1=0-1

z+1-1=-1

Finally
z= -1
i think thats how you do it

 

[tommykins]

please don't attempt 'complex numbers' when you don't know what it is.

 

[TBK11]

ah i thought i knew this soz

 

[vds700]

wouldn't it be

z^2+2z x 1+1^2=0

(z+1)^2=0

{ (z+1)^2} ^1/2=0

2x1/2
(z+1) =0

(z+1)^2/2=0

(z+1)^1/1=0

(z+1)^1=0

z+1=0

(z+1)+(-1)=0+(-1)

(z+1)-1=0-1

z+1-1=-1

Finally
z= -1
i think thats how you do it

its not a quadratic in z, so you cant solve it like that, Tommykins way is correct

you end up with equations
(x + 1)² - y² = 0
y = +(x + 1)

and
y(x - 1)=0 subbing in y
(x + 1)(x - 1)=0
x = +1

sub into y, u get y = +2
final solutions are (i think) 1 + 2i and 1 - 2i

 

[cutemouse]

wouldn't it be
....

Hahahah!!! |NOVICE|

 

[Affinity]

haven't looked carefully but surely z=-1 works.. so you must be missing something

 

[Timothy.Siu]

yeah he missed that solution

 

[vds700]

haven't looked carefully but surely z=-1 works.. so you must be missing something

ah yeah, that does work. I assumed that all the roots would be complex.

But how can this equation of degree 2 have 3 roots?

 

[Affinity]

As you said it's not a quadratic.

Something on the side.. It happens that if you consider real numbers only, then polynomials can appoximate any continuous function uniformly over a closed interval.

The analogous situation does not hold for polynomials over the complex numbers, but if you consider polynomials expressions with conjugates. then you can again approximate every functions over a "compact set".


So in a sense the complex conjugate is a lot different to ordinary functions

 

[cookiesandcream]

lol joining in the fun
x^2 + 2xyi - y^2 +2x -2yi + 1 =0
Equating reals:
x^2 -y^2 + 2x + 1 = 0 ------------------- [1]

Equating imaginary:
2xy - 2y = 0
i.e. 2y(x-1) = 0
y = 0 or x = 1

Sub y = 0 into equation [1]
x^2 +2x + 1 = 0
(x + 1) ^2 = 0
x = -1

Sub x = 1 into [1]
1 - y ^2 +2 +1 = 0
i.e. y= +/- 2

Solutions are : -1, 1 +/- 2 i