iiiiiiiiiiiii (1 Viewer)

[fullonoob]

Solve the equation 2z^4 + 3z^3 + 5z^2 + 3z +2 = 0
this iis what i have
dividing by 2 you get

2z^2 + 3z + 5 + 3z^-1 + 2z^-2 = 0
2(z^2 + z ^-2) + 3 (z + z^-1) + 5 = 0
Let u = z + z^-1
u^2 = (z + z^-1)^2
= z^2 + z^-2 +2
2(u^2-2) + 3u +5 = 0
2u^2 +3u+1 = 0
(2u +1)(u+1) = 0
2z +2z^-1 +1 = 0 OR z + z^-1 +1=0
2z^2 + z +2 = 0 OR z^2 + z +1 = 0 >>>>>>>THIS IS THE PART I DONT GET :angry:
I know you use u^2 = = z^2 + z^-2 +2, but if you root that it becomes all root
so im completely lost there...
Dont need to find the answer, just explain to me that step
Heelpppp please

 

[fullonoob]

someone answer already D:
bump

 

[addikaye03]

Solve the equation 2z^4 + 3z^3 + 5z^2 + 3z +2 = 0
this iis what i have
dividing by 2 you get

2z^2 + 3z + 5 + 3z^-1 + 2z^-2 = 0
2(z^2 + z ^-2) + 3 (z + z^-1) + 5 = 0
Let u = z + z^-1
u^2 = (z + z^-1)^2
= z^2 + z^-2 +2
2(u^2-2) + 3u +5 = 0
2u^2 +3u+1 = 0
(2u +1)(u+1) = 0
2z +2z^-1 +1 = 0 OR z + z^-1 +1=0
2z^2 + z +2 = 0 OR z^2 + z +1 = 0 >>>>>>>THIS IS THE PART I DONT GET :angry:
I know you use u^2 = = z^2 + z^-2 +2, but if you root that it becomes all root
so im completely lost there...
Dont need to find the answer, just explain to me that step
Heelpppp please

2z^4 + 3z^3 + 5z^2 + 3z +2 = 0

2z^2+3z+5+3z^-1+2z^-2=0

2(z^2+z^-2)+3(z+z^-1)+5=0

Let z=cos@+isin@, z+z^-1=(cos@+isin@)+(cos@+isin@)^-1

Using demoivres, RHS=(cos@+isin@)+cos(@)-isin(@)=2cos@

2(2cos2@)+3(2cos@)+5=0

10cos2@+6cos@+5=0

10(2cos^2@-1)+6cos@+5=0

20cos^2@+6cos@-5=0

cos@=[-6+-sqrt(436)]/40]

@=arccos(-6+-sqrt(436)]/40) where 0<=@<=2pi

EDIT:

2z^2 + z +2 = 0 OR z^2 + z +1 = 0 >>>>>>>THIS IS THE PART I DONT GET

I know you use u^2 = = z^2 + z^-2 +2, but if you root that it becomes all root

From here, you can just do what i did

 

[fullonoob]

jheeeebbus another question

3| z - (2+2i)| = |z-(6+6i)|
i) write down its radius and the coordinates of its centre

Can someone explain what the | | do
and how do you expand it please
Thanks very much
OMG FAIL

oh yeah and explain how the i disappears after you sub x+ iy

 

[fullonoob]

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sleeeeeeeeepyyyyyyyyy

 

[bachviete]

jheeeebbus another question

3| z - (2+2i)| = |z-(6+6i)|
i) write down its radius and the coordinates of its centre

Can someone explain what the | | do
and how do you expand it please
Thanks very much
OMG FAIL

oh yeah and explain how the i disappears after you sub x+ iy

The | | is the modulus, and we know



So



Also

 

[fullonoob]

The | | is the modulus, and we know



So



Also

You did it wrong at this step , you cant times the 3 in or it doesnt work out

But anyways i got it thanks


STUCK ON ANOTHER QUESTION NOW
i) Let OPQR be a square on an Argand diagram where O is the origin. The points P and R represent the complex numbers z and iz respectively. Find the complex number representing Q.
ii) The square is now rotated about O through 45* in an anticlockwise direction to OSTU. Find the complex numbers S, T and U.

i did i its easy.
P: z R: iz Q: (1+i)z
for ii) i didnt get the answer at the back
but i just times all of PRQ by cis 45...i think thats what you do o-o
help AGAIN LOL :uhhuh:

 

[jet]

Yes you do, which is the same as multiplication by (1/sqrt(2) + i/sqrt(2))

 

[cutemouse]

Solve the equation 2z^4 + 3z^3 + 5z^2 + 3z +2 = 0

I honestly don't think this is in the course, although Cambridge goes through it. But anyhow, another method:


and solve the quadratics, etc... I hope I haven't made any mistakes, hehe, LateX is hard to read when typing .

 

[Lukybear]

I honestly don't think this is in the course, although Cambridge goes through it. But anyhow, another method:


and solve the quadratics, etc... I hope I haven't made any mistakes, hehe, LateX is hard to read when typing .

How do you get from 4th last step to 3rd last step??

 

[cutemouse]

How do you get from 4th last step to 3rd last step??

Difference of two squares.