To solve complex number expressions like (√3 + i)/2 raised to high powers, use algebraic identities (a+b)² = a² + 2ab + b² and (a+b)³ = a³ + 3a²b + 3ab² + b³ to simplify the base, then leverage periodicity properties (x⁶ = -1) to reduce large exponents through division, ultimately finding that (√3 + i)/2^2027 = (√3 - i)/2.
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Germany | Can you solve this? | Math Olympiad
Added:Hello everyone, welcome to how to solve this very nice expression root 3 plus I divided by 2 whole power 2027.
So, let's start.
First, we suppose that let X is equal to root 3 plus I divided by 2.
It means that now we have to find the value of X to the power 20 27.
Now, we square both sides of this equation. This will become X squared equal to root 3 plus I divided by 2 whole squared.
Expand this numerator root 3 plus I by using this algebraic identity A plus B whole squared equal to A squared plus 2 AB plus B squared.
This will become X squared equal to root 3 squared plus 2 * root 3 * I plus I squared divided by 2 squared is uh 4.
Next, X squared equal to this square will be canceled out with this square root and here we are left with 3. Plus, this will become 2 * root 3 * I.
And this I squared I squared equal to -1.
So, this uh plus I squared will become -1 divided by four.
Further simplify, this will become x squared equal to 3 -1 will become two.
Plus two times two times root three I divided by four.
Next, x squared equal to From these two terms, we can factor out two. In bracket left, 1 plus root three times I divided by This four is same as two times uh two. So, this two will be canceled out with this two and we get the value of x squared equal to 1 plus root three I divided by two.
Now, we take a cube of both sides of uh this equation, x is equal to root three plus I over two.
We have x is equal to root three plus I over two. And we take uh cube of both sides.
So, this will become x cubed equal to Expand this root three plus I whole cubed by using this algebraic identity, a plus b whole cubed equal to a cubed plus three times a squared b plus three times a b squared plus b cubed.
So, this is root three plus i whole cubed will become root three whole cubed.
Plus three times root three squared times i plus three times root three times i squared plus i cubed.
Divided by two cubed is equal to eight.
Next, x cubed equal to root three cubed will become three times root three.
Plus this is squared will be cancelled out with this square root and three times three times i will become nine i.
And three times root three times i squared this i squared is equal to negative one.
So, three times root three times negative one will become negative three times root three.
Plus this i cubed can be written as i squared times i.
Divided by eight.
Next, x cubed equal to this negative three times root three will be cancelled out with this positive three times root three.
And in the numerator we are left with the nine times i since i squared is equal to negative one so negative one times i will become negative i.
Divided by eight.
Next, x cubed equal to 9i - 1i will become 8 * i / 8. So, this 8 will be canceled out and we are left with x cubed equal to i.
Now, we take a square of both sides.
So, x cubed whole square will become x to the power 6 equal to i squared.
Since i squared equal to -1, so x to the power 6 will be equal to 1.
And we have to calculate the value of x to the power 2027.
Since 2027 2027 is equal to 6 * 337 + 5.
So, we can rewrite this x to the power 2027 as x to the power 6 * 300 37 * x to the power 5.
This is equal to x to the power 2027.
And this x to the power 2027 can be written as x to the power 6 whole power 337 * x to the power 5. Now, we calculate the value of this x to the power 5.
To calculate the value of x to the power 5, we multiply this equation x cubed equal to i by this equation.
x squared equal to 1 + root 3 i over 2.
We have two equations x cubed equal to i and x squared equal to 1 + root 3 * i divided by 2. We multiply these two equations to get x to the power 5.
x to the power 5 is equal to x squared * x cubed.
So, this will become 1 + root 3 i divided by 2 times i.
So, 1 * i is i and root 3 * i * i will become + root 3 i squared divided by 2.
And because i squared equal to -1, so this will become x to the power 5 is equal to i minus root 3 divided by 2.
Now, because we have x to the power 5 equal to i minus root 3 divided by 2, so we replace this x to the power 5 with this uh i minus root 3 over 2.
And we replace this x to the power 6 with this negative 1.
So, this will become negative 1 to the power 337 * x to the power 5 is uh i minus root 3 divided by 2.
Since this uh 337 is an odd number, so this negative 1 remains negative 1.
This will become negative 1 * i minus root 3 divided by 2.
And uh negative 1 * i will become positive i, and this negative root 3 will become positive root 3 minus i divided by 2.
And this expression is equal to this uh x to the power 2027, x to the power 2027.
Since uh we have supposed that x to the power 2027 equal to this given expression.
So, the given expression root 3 plus i divided by 2 whole power 2027 is equal to root 3 minus I divided by 2.
This is the final answer of this problem.
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