Whats the answer? | Mathematics | Page 1

cold lynx Apr 19, 2024, 1:47 PM

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vestal ploverBOT Apr 19, 2024, 1:47 PM

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cold lynx

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somber fulcrum Apr 19, 2024, 1:47 PM

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Do you know the biquadratic root of 256?

cold lynx Apr 19, 2024, 1:47 PM

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yh

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idk what to do with the negative or is r=4i

somber fulcrum Apr 19, 2024, 1:48 PM

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cold lynx yh

What is it?

cold lynx Apr 19, 2024, 1:49 PM

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4

somber fulcrum Apr 19, 2024, 1:49 PM

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cold lynx 4

It's 16i.

cold lynx Apr 19, 2024, 1:49 PM

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how

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holy shit

somber fulcrum Apr 19, 2024, 1:50 PM

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cold lynx how

Sqrt(sqrt(-256))?

cold lynx Apr 19, 2024, 1:50 PM

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wait i think i see what i need to do

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alr holdd up

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nvm in tweaking

somber fulcrum Apr 19, 2024, 1:56 PM

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16i = rcis(theta).

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Find arg(16i).

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Do you know it?

cold lynx Apr 19, 2024, 1:57 PM

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wait so after r^4cis(4theta)=-256 then what

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wait now i got the asnwer

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its just 4cispie

somber fulcrum Apr 19, 2024, 2:03 PM

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cold lynx wait so after r^4cis(4theta)=-256 then what

Where did 4theta come from?

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Also z = 16i?

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z^4 = -256.

cold lynx Apr 19, 2024, 2:07 PM

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somber fulcrum Where did 4theta come from?

De Moivre’s

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im confused on biquadratic u dont need to do that

somber fulcrum Apr 19, 2024, 2:08 PM

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cold lynx im confused on biquadratic u dont need to do that

You do?

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How do you find z?

cold lynx Apr 19, 2024, 2:09 PM

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somber fulcrum How do you find z?

its in cis form

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u have to find it in cis form so u the theorem and expan so -256=r^4(cos4theta+isin4theta)

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and pie is -1,0

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so

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4cis(pie)=-256

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4cis(pie)=z^4 and then u just solve for z

somber fulcrum Apr 19, 2024, 2:12 PM

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Bro z^4 = -256.

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They said z = rcis(theta).

cold lynx Apr 19, 2024, 2:12 PM

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yh

somber fulcrum Apr 19, 2024, 2:12 PM

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Why square on both sides twice?

cold lynx Apr 19, 2024, 2:12 PM

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bro

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do you know polar form?

somber fulcrum Apr 19, 2024, 2:13 PM

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It's easier to do it normally.

somber fulcrum Apr 19, 2024, 2:13 PM

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cold lynx do you know polar form?

Yea.

cold lynx Apr 19, 2024, 2:14 PM

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somber fulcrum Yea.

so do u know de moivre's theorem?

somber fulcrum Apr 19, 2024, 2:14 PM

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Yea.

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Just r^n(cos(theta) + isin(theta)) that one?

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= z^n.

cold lynx Apr 19, 2024, 2:15 PM

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no it isnt

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r^n(cos(ntheta) + isin(ntheta))

somber fulcrum Apr 19, 2024, 2:15 PM

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somber fulcrum Just r^n(cos(theta) + isin(theta)) that one?

My mistake, I didnt multiply by theta.

cold lynx Apr 19, 2024, 2:16 PM

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alr 👍🏿

somber fulcrum Apr 19, 2024, 2:16 PM

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Anyway you gonna continue?

cold lynx Apr 19, 2024, 2:17 PM

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uh

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so its r^n(cos(theta) + isin(theta))

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=-256

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and thats 256x-1

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so 256=z^4

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and (cos(4theta) + isin(4theta))=-1

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and u get ur answer

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then sub 0-3 to get values

somber fulcrum Apr 19, 2024, 2:18 PM

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Oh ok.

cold lynx Apr 19, 2024, 2:18 PM

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yh

somber fulcrum Apr 19, 2024, 2:18 PM

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I was thinking of doing it by doing the biquadratic but this works too.

#Whats the answer?