complex stuff (1 Viewer)

[bob fossil]

Got an exam tommoro and im stuck on how to solve this.
z^2+(4-2i)z+6=0


I used quadratic form but not sure what to do with the square under the root because i can find the argument.

please help =]

 

[annabackwards]

Quadratic formula is the way to go.
z = -(4-2i) +/- sqrt( (4-2i)^2 -4x6) / 2
= -(4-2i) +/- sqrt( 16 - 4 -16i - 24) / 2
= -(4-2i) +/- sqrt( -12 -16i) / 2

Now to solve the squareroot, let a + ib = sqrt( -12 -16i)
Squaring both sides we get (a+ib)^2 =-12 -16i
So a^2 - b^2 + 2abi = -12 -16i
Equating real and imaginary parts we get
a^2 - b^2 = -12
a = -8/b <--------------------- sub this into above equation we get:
64/b^2 - b^2 = -12
64 - b^4 = - 12b^2
b^4 - 12b^2 - 64 = 0
(b^2 - 16)(b^2 + 4) = 0
b = +/- 4 (b must be real)
If b = +4, a = -2. If b = -4, a = +2

Thus sqrt( -12 -16i) = +/- (2 - 4i). Subbing this back into the quadratic formula we get:

z = -(4-2i) +/- +/- (2 - 4i) / 2
= -(4-2i) +/- (2 - 4i) / 2
= -(4-2i) + (2 - 4i) / 2 or = -(4-2i) - (2 - 4i) / 2

I'm sure you can do the rest

 

[bob fossil]

Quadratic formula is the way to go.
z = -(4-2i) +/- sqrt( (4-2i)^2 -4x6) / 2
= -(4-2i) +/- sqrt( 16 - 4 -16i - 24) / 2
= -(4-2i) +/- sqrt( -12 -16i) / 2

Now to solve the squareroot, let a + ib = sqrt( -12 -16i)
Squaring both sides we get (a+ib)^2 =-12 -16i
So a^2 - b^2 + 2abi = -12 -16i
Equating real and imaginary parts we get
a^2 - b^2 = -12
a = -8/b <--------------------- sub this into above equation we get:
64/b^2 - b^2 = -12
64 - b^4 = - 12b^2
b^4 - 12b^2 - 64 = 0
(b^2 - 16)(b^2 + 4) = 0
b = +/- 4 (b must be real)
If b = +4, a = -2. If b = -4, a = +2

Thus sqrt( -12 -16i) = +/- (2 - 4i). Subbing this back into the quadratic formula we get:

z = -(4-2i) +/- +/- (2 - 4i) / 2
= -(4-2i) +/- (2 - 4i) / 2
= -(4-2i) + (2 - 4i) / 2 or = -(4-2i) - (2 - 4i) / 2

I'm sure you can do the rest

Thank you so much =]


I'm to silly to see that I needed to expand the square lol

 

[annabackwards]

Thank you so much =]


I'm to silly to see that I needed to expand the square lol

No worries, i'm glad i helped

Good luck with your exam tomorrow!

 

[tommykins]

A quick way is using the modulus^2 as part of your simultaneous equations.

Sometimes saves you from having to solve quartics.

 

[annabackwards]

A quick way is using the modulus^2 as part of your simultaneous equations.

Sometimes saves you from having to solve quartics.

OOO i remember doing that once, but then that method never came up again so i forgot how to do it. Can you pretty please show me? :uhhuh:

 

[bob fossil]

Thank you hopefully the hard work pays off this time

 

[Lukybear]

A quick way is using the modulus^2 as part of your simultaneous equations.

Sometimes saves you from having to solve quartics.

Yeas how how?? Teach us!!

 

[bob fossil]

if i have z+1/z-1 and z =x+iy


what do i times the top and bottom by

 

[annabackwards]

if i have z+1/z-1 and z =x+iy


what do i times the top and bottom by

z+1/z-1 = x+1+iy/x-1+iy
You need to multiply the top and bottom by the denominator's conjugate, so multiply top and bottom by (x-1) - iy

 

[lyounamu]

Quadratic formula is the way to go.
z = -(4-2i) +/- sqrt( (4-2i)^2 -4x6) / 2
= -(4-2i) +/- sqrt( 16 - 4 -16i - 24) / 2
= -(4-2i) +/- sqrt( -12 -16i) / 2

Now to solve the squareroot, let a + ib = sqrt( -12 -16i)
Squaring both sides we get (a+ib)^2 =-12 -16i
So a^2 - b^2 + 2abi = -12 -16i
Equating real and imaginary parts we get
a^2 - b^2 = -12
a = -8/b <--------------------- sub this into above equation we get:
64/b^2 - b^2 = -12
64 - b^4 = - 12b^2
b^4 - 12b^2 - 64 = 0
(b^2 - 16)(b^2 + 4) = 0
b = +/- 4 (b must be real)
If b = +4, a = -2. If b = -4, a = +2

Thus sqrt( -12 -16i) = +/- (2 - 4i). Subbing this back into the quadratic formula we get:

z = -(4-2i) +/- +/- (2 - 4i) / 2
= -(4-2i) +/- (2 - 4i) / 2
= -(4-2i) + (2 - 4i) / 2 or = -(4-2i) - (2 - 4i) / 2

I'm sure you can do the rest

Instead of that

let (x+yi)^2 = -12 - 16i
x^2 + 2xyi - y^2 = -12-16i
x^2 - y^2 = -12 ...(1) and 2xy = -16
(x^2 + y^2)^2 = (x^2-y^2)^2 + 4x^2y^2
(x^2 + y^2)^2 = 144 + 256
x^2 + y^2 = 20 ...(2)

(1) + (2)
2x^2 = 8
x = +- 2
y = -+ 4

 

[bob fossil]

z+1/z-1 = x+1+iy/x-1+iy
You need to multiply the top and bottom by the denominator's conjugate, so multiply top and bottom by (x-1) - iy

Id rep you but i just did lol


thank you again =]

 

[Drongoski]

 

[cyl123]

Modulus method

(a+ib)^2=-12-16i
|(a+ib)^2|=|-12-16i|
noting |z^2|=|z|^2
|a+ib|^2=sqrt(144+256)
noting |a+ib|=sqrt(a^2+b^2)
a^2+b^2=20

Then combine with a^2-b^2 to get the answer

 

[lyounamu]

modulus method

(a+ib)^2=-12-16i
|(a+ib)^2|=|-12-16i|
noting |z^2|=|z|^2
|a+ib|^2=sqrt(144+256)
noting |a+ib|=sqrt(a^2+b^2)
a^2+b^2=20

then combine with a^2-b^2 to get the answer

+1

 

[annabackwards]

How do you use that method so seamlessly? Are there tricks or do you just have to practise?

Modulus method

(a+ib)^2=-12-16i
|(a+ib)^2|=|-12-16i|
noting |z^2|=|z|^2
|a+ib|^2=sqrt(144+256)
noting |a+ib|=sqrt(a^2+b^2)
a^2+b^2=20

Then combine with a^2-b^2 to get the answer

Oh i see, thank you! That is much easier - i'll remember it this time

 

[Drongoski]

How do you use that method so seamlessly? Are there tricks or do you just have to practise?


For the more obvious & simpler ones. Familiarity & practice helps I suppose.

What we were doing was to convert -12 - 16i into a perfect square.

If you have to convert -i into a perfect square it is rather inconvenient
If you have -2i you can convert via: -2i = 1 -2i - 1 = 1 -2i + (-i)² = (1-i)²
Therefore to find square root of -i, use i = 1/2 (-2i) and proceed.

 

[annabackwards]

For more obvious & simpler ones. Familiarity & practice helps I suppose.

Interesting. Thanks for replying

 

[Drongoski]

Interesting. Thanks for replying

Please see my edited reply.

 

[Lukybear]

How do you use that method so seamlessly? Are there tricks or do you just have to practise?

I think its from Terry Lee's book? Unless im mistaken.