Find the real part of (sqrt{3} - i)^{2011}.
let's use De Moivre's Theorem
let z = √3 - 1
sketching it in the Argand plane will give you
|r| = 2, and tanØ = -1/√3
Ø = 330° or 11π/6
z = 2(cos 330° - i sin 330°)
z^2011 = 2^2011(cos (663630°) - i sin(663630°) )
reducing 663630° to a standard angle by dividing multiples of 360°
663630° -----> 150°
z^2011 = 2^2011 (cos 150 + i sin 150°)
= 2^(-√3/2 + (1/2) i )
so the real part is (2^2011)(-√3/2)
= - 2^2010 √3
clearly I have a typo in the 3rd last line, should be
= 2^2011 (-√3/2 + (1/2) i )
To find the real part of the complex number (\sqrt{3} - i)^{2011}, we can use De Moivre's Theorem.
De Moivre's Theorem states that for any complex number z = r(cosθ + isinθ), where r is the magnitude and θ is the argument, the nth power of z can be written as:
z^n = r^n(cos(nθ) + isin(nθ))
In this case, z = \sqrt{3} - i, so the magnitude r and argument θ can be found as follows:
r = √(3^2 + (-1)^2) = √10
θ = arctan(-1/√3) = -π/6
Now, we can substitute these values into De Moivre's Theorem:
(\sqrt{3} - i)^{2011} = (√10)^{2011}(cos(-\frac{2011\pi}{6}) + isin(-\frac{2011\pi}{6}))
Since we're interested in the real part of this expression, we only need to evaluate the cosine term:
cos(-\frac{2011\pi}{6}) = cos(-335\pi) = cos(0) = 1
Therefore, the real part of (\sqrt{3} - i)^{2011} is (√10)^{2011} * 1 = (√10)^{2011}.
To find the real part of the expression \((\sqrt{3} - i)^{2011}\), we can first express the complex number \(\sqrt{3} - i\) in polar form.
Let's find the magnitude and argument of \(\sqrt{3} - i\):
Magnitude (r):
\[ r = \sqrt{(\text{real part})^2 + (\text{imaginary part})^2} = \sqrt{3^2 + (-1)^2} = 2 \]
Argument (θ):
\[ \theta = \arctan\left(\frac{\text{imaginary part}}{\text{real part}}\right) = \arctan\left(\frac{-1}{\sqrt{3}}\right) = -\frac{\pi}{6} \]
Now, let's express \(\sqrt{3} - i\) in polar form:
\[ \sqrt{3} - i = 2(\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6})) \]
Next, we'll raise it to the power of 2011 using De Moivre's theorem, which states that for any complex number \( z = r(\cos(\theta) + i\sin(\theta)) \),
\[ z^n = r^n(\cos(n\theta) + i\sin(n\theta)) \]
Plugging in the values, we have:
\[ (\sqrt{3} - i)^{2011} = (2(\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6})))^{2011} \]
\[ = 2^{2011}(\cos(2011 \cdot -\frac{\pi}{6}) + i\sin(2011 \cdot -\frac{\pi}{6})) \]
Simplifying further, we get:
\[ = 2^{2011}(\cos(-\frac{671 \pi}{2}) + i\sin(-\frac{671 \pi}{2})) \]
\[ = 2^{2011}(\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})) \]
\[ = 2^{2011}(0 + i \cdot 1) \]
\[ = 0 + 2^{2011}i \]
Therefore, the real part of \((\sqrt{3} - i)^{2011}\) is 0.