Why SDEHS papers so hard MX2 (2 Viewers)

[lolcti]

maybe stop complaining and learn how to do it

 

[kendricklamarlover101]

also note the other question to prove conjugate laws exist in division? If u had 9-12 questions like that, how would u fare in the 50 min time duration? That is the QUESTION.

just consider the conjugate of after realising it its literally not that hard of a problem as u say it is

 

[kendricklamarlover101]

bro there is not. who are u to say there is?

 

[MoonMonster196884]

just consider the conjugate of after realising it its literally not that hard of a problem as u say it is

it isn't but it is at the same time.

 

[kendricklamarlover101]

it isn't but it is at the same time.

Taking the conjugate we get



as required

 

[kendricklamarlover101]

the hardest part was literally just typing it on latex after doing it on paper

 

[MoonMonster196884]

Taking the conjugate we get



as required

your last step is wrong, you needa proofread. and it assumes what we were intending to prove.

 

[MoonMonster196884]

View attachment 41693

symbolab says o-wise

 

[MoonMonster196884]

@carrotsss do u all agree?

 

[carrotsss]

@carrotsss do u all agree?

there should be a plus on the numerator in his second last line rather than a minus in the middle, aside from that he’s right I think

 

[carrotsss]

nah but then he asms

the numerator* is x-iy and denom* is a+ib which we cant do bcwe havent proven that yet.

what are u waffling, he multiplied both sides by a-ib

 

[MoonMonster196884]

what are u waffling, he multiplied both sides by a-ib

heugh. his "as required" assumes that the conjugate of the numerator / denominator = conjugate of the numerator / conjugate of the denominator.
simple enough right? @carrotsss

 

[carrotsss]

heugh. his "as required" assumes that the conjugate of the numerator / denominator = conjugate of the numerator / conjugate of the denominator.
simple enough right?

that was due to an error in the second last line as I explained earlier

 

[MoonMonster196884]

that was due to an error in the second last line as I explained earlier

but who is he to say that the conjugate of z1/z2 is the conjugate of z1/ conjugate of z2 when he hasnt proven it yet?

 

[carrotsss]

but who is he to say that the conjugate of z1/z2 is the conjugate of z1/ conjugate of z2 when he hasnt proven it yet?

he didn’t say that he just made a typo in the second last line

 

[MoonMonster196884]

he didn’t say that he just made a typo in the second last line

this:

(but replace with a+ib) does not have to be the conjugate of

, does it..

 

[carrotsss]

this:

(but replace with a+ib) does not have to be the conjugate of

, does it..

im honestly lost atp ive forgotten mx2

 

[liamkk112]

this:

(but replace with a+ib) does not have to be the conjugate of

, does it..

ur right. using the rule that conj(z1z2) = conj(z1)conj(z2), then in this case u would end up with something of the form conj(1/a+bi)=(a-ib)/(a^2+b^2) which would have to then be multiplied by conj(x+iy) = x-iy, so overall conj(x+iy/a+bi) = (x-iy)(a-bi)/(a^2+b^2), i believe but im rusty

 

[liamkk112]

ur right. using the rule that conj(z1z2) = conj(z1)conj(z2), then in this case u would end up with something of the form conj(1/a+bi)=(a-ib)/(a^2+b^2) which would have to then be multiplied by conj(x+iy) = x-iy, so overall conj(x+iy/a+bi) = (x+iy)(a+bi)/(a^2+b^2), i believe but im rusty

that being said im pretty sure the equation u sent has solutions, if its still under discussion. as someone else already said u can just use the quadratic formula

 

[kendricklamarlover101]

ur right. using the rule that conj(z1z2) = conj(z1)conj(z2), then in this case u would end up with something of the form conj(1/a+bi)=(a-ib)/(a^2+b^2) which would have to then be multiplied by conj(x+iy) = x-iy, so overall conj(x+iy/a+bi) = (x-iy)(a-bi)/(a^2+b^2), i believe but im rusty

the conjugate of is not

so the conjugate is
so using this method u still get the same answer as i did as u get