calculate the value of complex?

Question

Evaluate: $$\left(\frac{−1 + i \sqrt3}{\sqrt2 + i\sqrt2}\right)^{20}$$

I got the value $-\frac{i}{\sqrt 2}$. Is it correct? My professor told me that I’m wrong....but I’m not getting how to calculate this questions.

Answer 1

Note that the expression can be written as: $$\frac{-1+i\sqrt 3}{\sqrt 2(1+i)} = \frac{-1 + i\sqrt 3}{2} \times \frac{\sqrt 2}{1+i}$$ $$=e^{\frac{2\pi i}{3}} \times \frac{1}{e^{\frac{i \pi}{4}}}$$ $$=e^{\dfrac{5\pi i}{12}}$$ using the well-known Euler’s formula.

Now, use De Moivre’s formula to conclude.

Answer 2

Hint:

Using How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i*\sin(\varphi)$?,

$$-1+i\sqrt3=\sqrt{(-1)^2+(\sqrt3)^2}\cdot e^{i2\pi/3}$$

$$\sqrt2(1+i)=\sqrt2\sqrt{1^2+1^2}\cdot e^{i\pi/4}$$

Answer 3

$$\begin{align} \frac{−1 + i \sqrt3}{\sqrt2 + i\sqrt2}&=\frac{−\frac12 + i \frac{\sqrt3}2}{\frac{\sqrt2}2 + i\frac{\sqrt2}2}\\&=\frac{e^{i\arctan(-\sqrt3)}}{e^{i\arctan(1)}}\\&=e^{i\Big(-\frac{\pi}3-\frac{\pi}4\Big)}\end{align}$$ Therefore, $$\Big(\frac{−1 + i \sqrt3}{\sqrt2 + i\sqrt2}\Big)^{20}=e^{20\times\Big(-\frac{7\pi i}{12}\Big)}$$

Answer 4

Note that $\omega=-1+i\sqrt 3=-1+\sqrt{-3}$ contains just one square root and will therefore be the root of a quadratic equation - you will be able to express $\omega^2$ as a linear expression in $\omega$ and thus compute powers quite efficiently.

So we do $\omega^2=(-1+i\sqrt 3)^2=-2-2i\sqrt 3=-2\omega -4$. $x=\omega$ is therefore one solution of the quadratic equation $x^2+2x+4=0$ (check that this works).

Now we can use this to compute higher powers so that $\omega^3=-2\omega^2-4\omega=8=2^3$. This is a helpful result, clearly (and could have been anticipated if you were familiar with the form of cube roots of $1$). From this the numerator comes out as $$\omega^{20}=\omega^{18}\omega^2=2^{18}\omega^2$$

Then also $(1+i)=1+\sqrt {-1}$ so we can use the same method. Here we find, even more helpfully, that $(1+i)^2=2i$ so that $(1+i)^{20}=(2i)^{10}$. Again, being familiar with the form of the roots of unity helps to spot that this will be a useful thing to do.

These two observations simplify the arithmetic hugely. I'll leave you to finish the calculations.