[UWSL]If z=(1+3√i/1−3√i) to the power of 10, then the value of 2 arcsen (Re(z)) + 5 arctg (2 Im(z)) is equal to[/UWSL]
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[UWSL][/UWSL]
IFUCKING HATE MATH AAAAAAAAAAAAAAAAAAAAAAAAAA
Why?IFUCKING HATE MATH AAAAAAAAAAAAAAAAAAAAAAAAAA
I DON'T UNDERSTAND THEM FUCK UUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU
Why you hate anime then?I DON'T UNDERSTAND THEM FUCK UUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU
JEWS
I don't have a choice, I think yes, fuck my hypocrisy I am a shame
(I like brunettes white pale skin so much, how did you know this?)
I mean he's not wrong
Do you mean z=(1+√3 i/1−√3i) to the power of 10
Yes
Are you asking these for geniune help of are you just trying to find the high iq cels
Both
Take k^10=Z
Therfore k=(1+√3 i/1−√3i)
You solve this by times both sides of the fraction by 1+√3 i [You can do this as this is the same as timsing by 1]
You get k=-1/2+1/2root3 i
You then use demoivres theorem as using normal binomail methods take too long
It states
(cos(theta)+isin(theta)^n=cos(ntheta)+isin(ntheta)
k=cos(2/3pi)+isin(2/3pi)
k^10=cos(20/3pi)+isin(20/3pi) [According to demovires theorem]
k^10=z so we found out what z is
z=-1/2+root3 /2 i
2*arcsin(-1/2) +5*arctan(root3/2 *2)=
-1/3pi+5/3pi=4/3pi
Therfore k=(1+√3 i/1−√3i)
You solve this by times both sides of the fraction by 1+√3 i [You can do this as this is the same as timsing by 1]
You get k=-1/2+1/2root3 i
You then use demoivres theorem as using normal binomail methods take too long
It states
(cos(theta)+isin(theta)^n=cos(ntheta)+isin(ntheta)
k=cos(2/3pi)+isin(2/3pi)
k^10=cos(20/3pi)+isin(20/3pi) [According to demovires theorem]
k^10=z so we found out what z is
z=-1/2+root3 /2 i
2*arcsin(-1/2) +5*arctan(root3/2 *2)=
-1/3pi+5/3pi=4/3pi
The answer is, you need a new hobby
Correct!Take k^10=Z
Therfore k=(1+√3 i/1−√3i)
You solve this by times both sides of the fraction by 1+√3 i [You can do this as this is the same as timsing by 1]
You get k=-1/2+1/2root3 i
You then use demoivres theorem as using normal binomail methods take too long
It states
(cos(theta)+isin(theta)^n=cos(ntheta)+isin(ntheta)
k=cos(2/3pi)+isin(2/3pi)
k^10=cos(20/3pi)+isin(20/3pi) [According to demovires theorem]
k^10=z so we found out what z is
z=-1/2+root3 /2 i
2*arcsin(-1/2) +5*arctan(root3/2 *2)=
-1/3pi+5/3pi=4/3pi
Are you a STEMcel?
What hobbies do you have but browsing forums all day
Does banging your mom count?
Im a mathcel so yh
Aren't you supposed to be dead you made a whole thread talking about you killing yourself
but apparently your too much of a bitch to do that right
Your mom called me after reading the thread and promised me a blowjob if i don’t.Aren't you supposed to be dead you made a whole thread talking about you killing yourself
You owe me money
Do you not know the answer but not the method, which is why you ask these problems
I could solve this problem, I ask those to find high IQcels to solve problems I can't, such as this one:
What's the solution to this system?
4pi/3?
Glad this clarification was made. Its a lot less tedious now.
Idk math
Correct!
First part: [Sorry if the formatting is bad]
So through various log laws you should be able to get this:
log3logroot3(x)-logroot3log3(y)=1
log32log3(x)-1/2log3log3(y)=1
2log3log3(x)-1/2log3log3(y)=1
This gives us
log3(log3(x))-1/2log3(log3(y)=1-log3(2)
3 to the power both sides
3^[log3(log3(x))-1/2log3(log3(y)]=3^[1-log3(2)]
3^[log3(log3(x))]*3^[-1/2log3(log3(y))]
log3(x)*1/sqrt(log3(y))
3^[1-log3(2)]=3*3^[-log3(2)]=1.5
So we can write this equation as:
log3(x)=1.5*sqrt(log3(y))
3 to the power both sides again
3^[log3(x)]=3^[1.5*sqrt(log3(y))]
x=(3^1.5)^sqrt(log3(y))
With x and y both being larger then 1
I checked with desmos this should be right
I think there might be something wrong with it but its my best guess for now
So through various log laws you should be able to get this:
log3logroot3(x)-logroot3log3(y)=1
log32log3(x)-1/2log3log3(y)=1
2log3log3(x)-1/2log3log3(y)=1
This gives us
log3(log3(x))-1/2log3(log3(y)=1-log3(2)
3 to the power both sides
3^[log3(log3(x))-1/2log3(log3(y)]=3^[1-log3(2)]
3^[log3(log3(x))]*3^[-1/2log3(log3(y))]
log3(x)*1/sqrt(log3(y))
3^[1-log3(2)]=3*3^[-log3(2)]=1.5
So we can write this equation as:
log3(x)=1.5*sqrt(log3(y))
3 to the power both sides again
3^[log3(x)]=3^[1.5*sqrt(log3(y))]
x=(3^1.5)^sqrt(log3(y))
With x and y both being larger then 1
I checked with desmos this should be right
I think there might be something wrong with it but its my best guess for now
Last edited:
I made a mistake
What's your IQ bhai?
It should be
log3(log3(x))-2log3(log3(y)=1-log3(2)
instead of...
log3(log3(x))-1/2log3(log3(y)=1-log3(2)
I believe all other steps are correct
Idk. Never got tested by professional
Where would you put yourself? Given that you can crack the JEE
x=3^(1.5log3(y)^2)
The other steps were all correct it was just that one mistake
You can verify this with desmos it works!
The other steps were all correct it was just that one mistake
You can verify this with desmos it works!
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