is there as shorter method for this?? am i being dumb (1 Viewer)

v.tex · 2025-01-21T11:51:33+1100

Screenshot 2025-01-21 at 11.50.02 AM.png

DO you actually have to expand it the whole way out just to find a??

pl4smaa21 · 2025-01-21T11:59:32+1100

I didn't check this but maybe you can find 1+sqrt(2) in modulus-argument form and then use de-moivre's theorem to more easily express z^4 , z^3 and z.

pl4smaa21 · 2025-01-21T12:58:16+1100

there I solved it that way it seems to work nicely

This method avoids the need to expand out unnecessarily as De Moivre's theorem allows you to express the powers of a complex number more succinctly (step 3). It's like a shortcut.

tywebb · 2025-01-21T16:57:08+1100

pl4smaa21 · 2025-01-21T16:58:45+1100

tywebb said:
$\begin{aligned}a&=\textstyle2z^2-z^3+\frac{9}{z}\\ &=\textstyle z^2(2-z)+\frac{9\bar z}{|z|^2}\\ &=\textstyle(-1+2\sqrt2 i)(1-\sqrt2 i)+\frac{9(1-\sqrt2 i)}{3}\\ &=3+3\sqrt2 i+3-3\sqrt2 i\\ &=6 \end{aligned}$

istg i'm retarded Lol that's a waayyyyy easier way to do it my sol. is such a drag lmao

liamkk112 · 2025-01-21T17:26:30+1100

v.tex said:
View attachment 46320

DO you actually have to expand it the whole way out just to find a??

Let $x = 1+ \sqrt{2}i$
$P(x) = 0 = x^4 - 2x^3 + ax - 9$
$\implies a = \frac{9+2x^3-x^4}{x}$
put that in your calculator and thats it (makes your life easier if you save 1+sqrt(2)i in your answer memory)

pl4smaa21 · 2025-01-21T17:28:09+1100

liamkk112 said:
Let $x = 1+ \sqrt{2}i$
$P(x) = 0 = x^4 - 2x^3 + ax - 9$
$\implies a = \frac{9+2x^3-x^4}{x}$
put that in your calculator and thats it (makes your life easier if you save 1+sqrt(2)i in your answer memory)

but ur calc needs that 'cmplx' function doesn't it? but yea ur right that's a good way

tywebb · 2025-01-21T19:36:10+1100

pl4smaa21 said:
istg i'm retarded Lol that's a waayyyyy easier way to do it my sol. is such a drag lmao

nah. u solved it before i did. so u not retarded.

pl4smaa21 · 2025-01-21T19:44:20+1100

tywebb said:
nah. u solved it before i did. so u not retarded.

thnx man.

I was just a bit shocked because of how straightforward you're solution was it shocked me that's why I wrote what I did out of emotion not seriousness so it's all good.

y'know when you can't solve a certain problem then you look at the solution and it all just comes together and makes like PERFECT SENSE and you just think to yourself "How did I not see that, it's so obvious!" that's how I felt basically

I hear that the above feeling of epiphany is rather common in STEM

tywebb · 2025-01-21T20:27:37+1100

well i've been doing this a long time so one should not be so shocked when some bro like me comes along and makes a more efficient solution

v.tex · 2025-01-21T20:33:15+1100

tywebb · 2025-01-21T20:38:45+1100

v.tex said:
View attachment 46333

$\begin{aligned}4z^3-iz^2-4z+i&=z^2(4z-i)-1(4z-i)\ \&=(z^2-1)(4z-i)\\ &=(z-1)(z+1)(4z-i) \end{aligned}$

pl4smaa21 · 2025-01-21T20:54:34+1100

tywebb said:
well i've been doing this a long time so one should not be so shocked when some bro like me comes along and makes a more efficient solution

oh no I didn't mean it like that like I am shocked that YOU in particular made a more efficient solution (i'm not that egotistical lol) I by no means expected my way to be the best moreso I was just surprised at straightforwardness of ur way like it felt natural y'know like that is what I felt SHOULD have came to mind first because of its simplicity but it didn't I instead went with the most dragged method using de moivre's lmao

tywebb · 2025-01-21T21:01:04+1100

yeah but i think it would not be such a surprise for someone with as much experience as me to come up with such a solution.

i would not necessarily expect the less experienced bros to come up with such a solution.

pl4smaa21 · 2025-01-21T21:16:13+1100

tywebb said:
yeah but i think it would not be such a surprise for someone with as much experience as me to come up with such a solution.

i would not necessarily expect the less experienced bros to come up with such a solution.

tru tru

liamkk112 · 2025-01-22T15:53:22+1100

pl4smaa21 said:
but ur calc needs that 'cmplx' function doesn't it? but yea ur right that's a good way

yes, but tbh its probably worthwhile to get the calculator with it lol. it saves a lot of time, especially in problems like these where you'd otherwise have to do a bunch of algebra, or to double check your answer quickly eg when finding roots of a polynomial. obviously you still have to show working, but this can help a lot

pl4smaa21 · 2025-01-22T15:58:14+1100

liamkk112 said:
yes, but tbh its probably worthwhile to get the calculator with it lol. it saves a lot of time, especially in problems like these where you'd otherwise have to do a bunch of algebra, or to double check your answer quickly eg when finding roots of a polynomial. obviously you still have to show working, but this can help a lot

yea tru I had a casio calc before that had the 'cmplx' and the 'vector' functions that would probs be useful for ext. maths in the future (i'm gonna begin y11 this year ) but my sister took it and that's why I said what I said above cuz I noticed there's the need for that kinda calc in ur solution. But yea ur right.

C_master · 2025-01-22T19:14:30+1100

pl4smaa21 said:
tru tru

why would u let urself be clocked by tywebb

pl4smaa21 · 2025-01-22T19:22:59+1100

C_master said:
why would u let urself be clocked by tywebb

idk man I'm a 2026'er lol

tywebb · 2025-01-22T20:15:08+1100

C_master said:
why would u let urself be clocked by tywebb

no clocking or any kind of offence intended

all i meant was that more seasoned veterans may be more likely to come up with a more efficient solution

i have every confidence that with a bit more experience pl4smaa21 will be able to do that too

is there as shorter method for this?? am i being dumb (1 Viewer)

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