Question

Show that (2 + √2i)^20+ (2 − √2i)^20is an integer.

Answer 1

Answer:

Yes, it is, we need to use the Moivre theorem and we get

Step-by-step explanation:

Hi, first, let´s introduce Moivre theorem to find the nth power of a complex number.

$z^{n} =r(Cos(n\alpha )+iSin(n\alpha ))$

Where:

r = module of the complex number

n= power

alpha= inclination angle

to find the module of the complex number, we need to use the following formula.

$r=\sqrt{a^{2} +b^{2} }$

Where:

z= a+bi

a= real part of the coplex number

b=imaginary part of the complex number

Finally, in order to find the angle (alpha), we have to use the following.

$\alpha =tan^{-1} (\frac{b}{a} )$

But, using Moivre for a complex number to the 20th power is not very practical, so we are going to assume some things first

$z_{1} =(2+\sqrt{2} i)$

$z_{2} =(2-\sqrt{2} i)$

So, first we are going to find the value of $z_{1} ^{2}$ and elevate it to the 10th power in order to get $(2+\sqrt{2} i)^{20}$

First, lets find the module of z1

$r_{1} =\sqrt{2^{2} +(\sqrt{2} )^{2} }=\sqrt{4+2} =\sqrt{6}$

and its angle is:

$\alpha =tan^{-1} (\frac{\sqrt{2} }{2} )=45$

we are all set, now let´s find the value of z_{1} ^{2}

$z_{1} ^{2} =\sqrt{6} (Cos(2*45 )+iSin(2*45 ))$

$z_{1} ^{2} =\sqrt{6} (Cos(90 )+iSin(90 ))$}

$z_{1} ^{2} =\sqrt{6} (0+i(1))$

$z_{1} ^{2}=\sqrt{6} i$

Now, let´s find the value of $z_{1} ^{20}$

$(z_{1} ^{2})^{10} =(\sqrt{6} i)^{10} =7,776(i)^{4} (i)^{4} (i)^{2} =7,776(1)(1)(-1)=-7,776$

therefore:

$(2 + \sqrt{2} i)^{20} =-7,776$

We do the same for (2 − √2i)^20, this time:

$z_{2} =(2-\sqrt{2})$

$r_{2} =\sqrt{2^{2} +(-\sqrt{2} )^{2} }=\sqrt{4+2} =\sqrt{6}$

And the angle is

$\alpha =tan^{-1} (\frac{-\sqrt{2} }{2} )=-45$

Therefore, we get:

$z_{2} ^{2} =\sqrt{6} (Cos(2*(-45) )+iSin(2*(-45) ))$

$z_{2} ^{2} =\sqrt{6} (Cos(-90 )+iSin(-90 ))$

$z_{2} ^{2} =\sqrt{6} (0+i(-1))$

$z_{2} ^{2}=-\sqrt{6} i$

Now, let´s find the value of $z_{2} ^{20}$

$(z_{2} ^{2})^{10} =(-\sqrt{6} i)^{10} =7,776(i)^{4} (i)^{4} (i)^{2} =7,776(1)(1)(-1)=-7,776$

therefore:

$(2 - \sqrt{2} i)^{20} =-7,776$

And then, we add them up

$(2+\sqrt{2} i)^{20}+(2-\sqrt{2} i)^{20}=-7,776+(-7,776)=-15,552$

So, yes, the result is an integer, -15,552