how is this the answer? (1 Viewer)

[kkk579]

 

[Average Boreduser]

View attachment 43657

hav u learnt vectors yet? your answer cant be purely imaginary bc your vector clearly has an acute direction

 

[kkk579]

hav u learnt vectors yet? your answer cant be purely imaginary bc your vector clearly has an acute direction

Shit ur right yeah i have learned vectors lemme see whwre i went wrong

 

[Average Boreduser]

imo use cis form for this

 

[kkk579]

imo use cis form for this

Hmmm where did i go wrong

 

[kkk579]

Hmmm where did i go wrong

I genuinely have no clue

 

[Average Boreduser]

hold on ill quickly see if i can pull it out lol (btw once u get the answer in vector form, u needa convert into eulers form)

 

[kkk579]

OHH NVM I SEE IT NOW

 

[kkk579]

OHH NVM I SEE IT NOW

Ik whwre i went wrong it was silly mistake LMAO

 

[kkk579]

wait shoulsnt the answer be 2e^i(2pi/3 + theta) not e^i x (theta plus 2pi/3)

 

[t234]

hey can i ask when ngo started mx2?? has it been a long time or

 

[kkk579]

hey can i ask when ngo started mx2?? has it been a long time or

Its been around 2.5 months

 

[t234]

is that the specialist class

Its been around 2.5 months

 

[kkk579]

this is what i did

 

[kkk579]

is that the specialist class

Yeah

 

[liamkk112]

View attachment 43657

instantly we can write z_2 = sqrt(3)iz_1
so z_2 -z_1 = z_1(-1+sqrt(3)i)
now z_1 = e^(ix) [im using x instead of theta]
and -1+sqrt(3)i = 2e^i(2pi/3) by conversion to modulus-argument form
thus, z_2 -z_1 = 2e^i(x+2pi/3)

 

[kkk579]

instantly we can write z_2 = sqrt(3)iz_1
so z_2 -z_1 = z_1(-1+sqrt(3)i)
now z_1 = e^(ix) [im using x instead of theta]
and -1+sqrt(3)i = 2e^i(2pi/3) by conversion to modulus-argument form
thus, z_2 -z_1 = 2e^i(x+2pi/3)

Yeah thats what i got in my new answer cos i realised i made a silly mistake in my original one but the solutions said something else