Question

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Answer 1

This is an odd question. What does the degree argument have to do with anything?

Answer 2

when it's in the form of r(cos(theta) + isin(theta)) the theta has to be between 90 and 180, since there'll be more than one solution ig idk

Answer 3

Okiday den. Lemme put a guess into the desmos and then I'll get back to ya.

Answer 4

.. it isn't a desmos thing

Answer 5

Ik ik. I just use desmos to do stuffs. Also, are only numbers allowed in those boxes?

Answer 6

yes

Answer 7

Sorry, I am completely lost on this one. I am unsure how to solve this.

Answer 8

@tranquility @AZ @florisalreadytaken @dude
when you guys become alive e.e please and thank you

Answer 9

@smokeybrown

Answer 10

I'm not very familiar with complex numbers, so I used this article to refresh a bit.
https://byjus.com/maths/argument-of-complex-numbers/
Apparently, the argument of a complex number is calculated using the formula
arg (z) = tan^-1(y/x)
This seems like it would still leave many possible answers that could be correct (even given the range between 90 and 180 degrees?), so I'm a bit confused on how to find the solution. I feel like I might not be approaching this the right way

Answer 11

https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:complex/x9e81a4f98389efdf:complex-mul-div-polar/a/complex-number-polar-form-review this should help you

Answer 12

i wouldnt even get into \( a+bi \) at all -- time consuming

you jcan just change it into:
\[ z^3=27 \ \ \Rightarrow \ \ z^3-3^3=0 \]
as you might know this formula:
\[ a^3-b^2=(a-b)(a^2+ab+b^2) \] where in our case \( a \) is z right? and \( b \) would be 3, the base of it

from that you would get two separate equations, respectively:
\[ (a-b)=0 \ \ \ and \ \ \ (a^2+ab+b^2) =0 \]
and that would give you the three solutions, exactly the same as the ones the link above says -- this is just an alternative way

and i think you did post one like this not long ago did you?

Answer 13

\[(z-3)(z^2+3z+9)=0\\z=3\] \[ \frac{-3 \pm \sqrt{9-4(9)}}{2} \\\frac{-3 \pm \sqrt{-27}}{2}\\\frac{-3 \pm 3i\sqrt{3}}{2} \]

\[z=3\\z=-1.5 + 1.5i\sqrt{3} \\ z=-1.5 - 1.5 i \sqrt{3}\]

Answer 14

but then later i have to get it to the form tho of a + bi

Answer 15

oh well,, i did not see the way the answer was supposed to be

to make it easier for you, \( 27\) is exactly the same as \( 27+0i \)

thus, we write:
\[ z^3=\left(\left(27+0i\right)^{\frac{1}{3}}\right)^3 \]
can you solve that?

Answer 16

uh dont the 1/3 and 3 make it 1... and then it'd be still 27 e.e

Answer 17

nah it would be 3+0i

Then just continue as you would normally do

Refer to what surjit sent you

Just math from there, really

Answer 18

You already know that a + bi is a complex number

When you write it in polar form, it would be in the form of \(r(\cos \theta + i \sin \theta)\)

so we can say that
z = a + bi

and another form is
\( z = r(\cos \theta + i \sin \theta)\)

so if you had \(z^n\)

that would be
\(\Large z^n = r^n[\cos(n \cdot \theta) + i \sin (\cdot \theta)]\)

and so remember how I said another way to write 'z' is as a + bi

so basically z^n is (a+bi)^n
and so we're trying to find that

now

z^3 = 27

27 is a real number, there's no imaginary number here
so as Flor said, you can re-write that as 27 + 0i

and so remember that formula
r^n (cos theta + i*sin theta)

But first does that make sense so far?

Answer 19

From https://courses.lumenlearning.com/precalctwo/chapter/polar-form-of-complex-numbers/

Answer 20

that's one over n tho o-O

Answer 21

wouldn't we just use what you wrote earlier o-0 \[z^n = r^n[\cos(n \cdot \theta) + i \sin (n\cdot \theta)]\]

Answer 22

but how do you use that to get the r and theta >.>

Answer 23

theta is arctan of b/a if some rule is true — i really cannot remember it at the moment ; just google it, and you will find it

and for r look above rah

Answer 24

3 + 0i, b is 0 though, it would be undefined

Answer 25

What? No — its 0

Answer 26

oh o:
arctan 0 would be 0 o:

Answer 27

🤝👏🏻👍🏻 — now find where theta belongs (i did this but someway something’s missing)

Answer 28

if it's 0 isn't theta 0 ._.

Answer 29

where it belongs you donnie

Answer 30

1. what is a donnie
2. i don't understand -.-

Answer 31

Then stay confused

One thing i dont get though is it why doesnit say to round it? I get real whole numbers as in the answer

Answer 32

maybe something different in radians and degrees o:

Answer 33

who does degrees in this?

And just fyi 90 degrees is there at the upper half of y axis, and 180 is at the right hand side of the x axis

so out number is in between there

But i am just confusing myself for some reason

Answer 34

yikesiez

Answer 35

would be intrigued if i am missing something basic — good luck AZ.

Answer 36

and you just leave- :|

Answer 37

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Answer 38

since we have three roots

z^3 means there are three roots
just like if you had a quadratic equation (x^2) then you would have two roots

so since we have three roots, you have to plug in k = 0, 1, 2 to find the different solutions

Answer 39

\[r^3(cos(120k) + i sin(120k) \]
?

Answer 40

your angle would be 360

because cos 360 is 1
and sin 360 gives you the 0

and so 360/n is inside the parenthesis
n is 3 because z^3

Answer 41

yeah

Answer 42

\[27(cos(120k) + i sin(120k)\]

Answer 43

wait no the angle inside the parenthesis for the polar form should still be 360

but when you're solving it, using that formula in the last screenshot I sent- the angle (theta) is 360 and then you divide by n

Answer 44

360/3 is 120-

Answer 45

\( z= 27(\cos(360k) + i \sin(360k)\)

Answer 46

\[27^{\frac{1}{3}}[cos(\frac{360}{3} + \frac{2k\pi}{3}) + i~sin(\frac{360}{3} + \frac{2k\pi}{3}) \\ 3[cos(120 + \frac{2k\pi}{3}) + i~sin(120+\frac{2k\pi}{3})\]
\[k = 0 \\ 3[cos(120 + \frac{2(0)\pi}{3}) + i~sin(120+\frac{2(0)\pi}{3}) \\3[cos(120) + i~sin(120)]\\ -\frac{3}{2} + \frac{3i \sqrt{3}}{2} \] using mathway


\[k=1 \\ 3[cos(120 + \frac{2\pi}{3}) + i~sin(120+\frac{2\pi}{3}) \\ \] It WoNt LeT mE sImPliFy ittttttttttttt


\[k=2 \\ 3[cos(120 + \frac{2(2)\pi}{3}) + i~sin(120+\frac{2(2)\pi}{3}) \\3[cos(120 + \frac{4\pi}{3}) + i~sin(120+\frac{4\pi}{3}) \]




although I did get the 3 roots earlier though,
\[z=3\\z=-1.5 + 1.5i\sqrt{3} \\ z=-1.5 - 1.5 i \sqrt{3} \] idkidk

Answer 47

Okay so the thing is 360 is degrees so you could just write that as 2pi lol and then you'll be able to simplify it


but you see that when k = 0 that's 120 degrees inside of cos and sin

that's going to be within 90 and 180 degrees and is your answer

Answer 48

then it would simplify to\[3[cos(240) + isin(240)]\]

Answer 49

and then be the root o-0

Answer 50

nvmd i finally figured it out *big dumb*

Answer 51

yes haha