Answer :
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.
[tex]z^{n} =r(Cos(n\alpha )+iSin(n\alpha ))[/tex]
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.
[tex]r=\sqrt{a^{2} +b^{2} }[/tex]
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.
[tex]\alpha =tan^{-1} (\frac{b}{a} )[/tex]
But, using Moivre for a complex number to the 20th power is not very practical, so we are going to assume some things first
[tex]z_{1} =(2+\sqrt{2} i)[/tex]
[tex]z_{2} =(2-\sqrt{2} i)[/tex]
So, first we are going to find the value of [tex]z_{1} ^{2}[/tex] and elevate it to the 10th power in order to get [tex](2+\sqrt{2} i)^{20}[/tex]
First, lets find the module of z1
[tex]r_{1} =\sqrt{2^{2} +(\sqrt{2} )^{2} }=\sqrt{4+2} =\sqrt{6}[/tex]
and its angle is:
[tex]\alpha =tan^{-1} (\frac{\sqrt{2} }{2} )=45[/tex]
we are all set, now let´s find the value of z_{1} ^{2}
[tex]z_{1} ^{2} =\sqrt{6} (Cos(2*45 )+iSin(2*45 ))[/tex]
[tex]z_{1} ^{2} =\sqrt{6} (Cos(90 )+iSin(90 ))[/tex]}
[tex]z_{1} ^{2} =\sqrt{6} (0+i(1))[/tex]
[tex]z_{1} ^{2}=\sqrt{6} i[/tex]
Now, let´s find the value of [tex]z_{1} ^{20}[/tex]
[tex](z_{1} ^{2})^{10} =(\sqrt{6} i)^{10} =7,776(i)^{4} (i)^{4} (i)^{2} =7,776(1)(1)(-1)=-7,776[/tex]
therefore:
[tex](2 + \sqrt{2} i)^{20} =-7,776[/tex]
We do the same for (2 − √2i)^20, this time:
[tex]z_{2} =(2-\sqrt{2})[/tex]
[tex]r_{2} =\sqrt{2^{2} +(-\sqrt{2} )^{2} }=\sqrt{4+2} =\sqrt{6}[/tex]
And the angle is
[tex]\alpha =tan^{-1} (\frac{-\sqrt{2} }{2} )=-45[/tex]
Therefore, we get:
[tex]z_{2} ^{2} =\sqrt{6} (Cos(2*(-45) )+iSin(2*(-45) ))[/tex]
[tex]z_{2} ^{2} =\sqrt{6} (Cos(-90 )+iSin(-90 ))[/tex]
[tex]z_{2} ^{2} =\sqrt{6} (0+i(-1))[/tex]
[tex]z_{2} ^{2}=-\sqrt{6} i[/tex]
Now, let´s find the value of [tex]z_{2} ^{20}[/tex]
[tex](z_{2} ^{2})^{10} =(-\sqrt{6} i)^{10} =7,776(i)^{4} (i)^{4} (i)^{2} =7,776(1)(1)(-1)=-7,776[/tex]
therefore:
[tex](2 - \sqrt{2} i)^{20} =-7,776[/tex]
And then, we add them up
[tex](2+\sqrt{2} i)^{20}+(2-\sqrt{2} i)^{20}=-7,776+(-7,776)=-15,552[/tex]
So, yes, the result is an integer, -15,552