Hi, can anyone solve this problem:
Determine z^n and all values of sqrts with base n of z.
z=(2-2i)^22 - (-1+isqrt(3))^33 , n=4
Why is this equality valid 2-2i = 2*sqrt(2)*e^-i*pi/4 = 2^3/2*e^-i*pi/4 ?
+118609
Determine z^n and all values of sqrts with base n of z.
z=(2-2i)^22 - (-1+isqrt(3))^33 , n=4
\(2-2i\\ = 2(1-i)\\ =2\sqrt2\left(\frac{1}{\sqrt2}+\frac{-i}{\sqrt2}\right)\\ =2\sqrt2\left(cos(\frac{-\pi}{4}+isin\frac{-\pi}{4}) \right)\\ =2\sqrt2e^{i*\frac{-\pi}{4}}\\ =2^{3/2}e^{-\pi/4}\)
\((2-2i)^{22}=\left[ 2^{3/2}e^{-(\pi /4)i} \right]^{22}\\ = 2^{33}e^{-(22\pi /4)i}\\ = 2^{33}e^{-(11\pi /2)i}\\ = 2^{33}e^{(-11\pi /2+12\pi/2)i}\qquad \mbox{I added 3 revolutions}\\ = 2^{33}e^{(\pi/2)i}\\ =2^{33}*i\)
\(- (-1+i\sqrt3)^{33}\\ =-(-1)^{33}(1-i\sqrt3)^{33}\\ =+(1-i\sqrt3)^{33}\\ =(\sqrt4)^{33}\left(\frac{1-i\sqrt3}{\sqrt{4}}{}\right)^{33}\\ =2^{33}\left(\frac{1-i\sqrt3}{2}{}\right)^{33}\\ =2^{33} \left[cos(-\pi/3)+isin((-\pi/3)i)\right]^{33}\\ =2^{33}\left[ e^{-\pi/3*i} \right]^{33}\\ =2^{33} e^{-11\pi *i} \\ =2^{33} e^{-11\pi *i+12\pi i} \\ =2^{33} e^{\pi i} \\ \)
\(=2^{33}*-1\\ =\;-2^{33}\)
\(z=2^{33}\;i-2^{33}\\ z=2^{33}\;(i-1)\\ ...\\ z^n=2^{33n}\;(i-1)^n\\ ....\\ z^4=2^{33*4}\;(i-1)^4\\ z^4=2^{132}\;(-1-2i+1)^2\\ z^4=2^{132}\;(-2i)^2\\ z^4=2^{132}\;*(-4)\\ z^4=\;-2^{132}\;*2^2\\ z^4=\;-2^{134}\\ z^4\approx \;-2.1778\times 10^{40} \)
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(I did the second part first)
Why is this equality valid
2-2i = 2*sqrt(2)*e^-i*pi/4 = 2^3/2*e^-i*pi/4 ?
I will add some brackets here. You should have added the brackets yourself!
2-2i = 2*sqrt(2)*e^(-i*pi/4) = 2^(3/2)*e^(-i*pi/4) ? I assume that is what you want?
Firstly
\(2^{3/2}=2^{1+1/2}=2^1*2^{1/2}=2\sqrt2\)
so 2*sqrt(2)*e^(-i*pi/4) = 2^3/2*e^(-i*pi/4)
Now to show that they are the same as the first part.
i.e. show that
2-2i = 2*sqrt(2)*e^(-i*pi/4)
\(RHS=2*\sqrt2*e^{(-i*\pi /4)}\\ RHS=2*\sqrt2*e^{(-i*\pi /4)}\\ RHS=2*\sqrt2*[cos(-\pi /4)+isin(-\pi /4)]\\ RHS=2*\sqrt2*\left[\frac{1}{\sqrt2}+i*\frac{-1}{\sqrt2}\right]\\ RHS=2*\sqrt2*\left[\frac{1}{\sqrt2}-\frac{i}{\sqrt2}\right]\\ RHS=2(1-i)\\ RHS=2-2i\\ RHS=LHS \)
z=(2-2i)^22 - (-1+isqrt(3))^33 , n=4 =-2.17780 × 10^40
Why is this equality valid 2-2i = 2*sqrt(2)*e^-i*pi/4 = 2^3/2*e^-i*pi/4 ?
This appears to be FALSE!.
Not sure what you are asking in the first part.
The equality is correct.
Remember that
\(e^{\imath\theta}= \cos\theta + \imath \sin\theta \)
and that
\(\cos(-\pi/4)=1/\sqrt{2}\),
\(\sin(-\pi/4)=-1/\sqrt{2}.\)
+118609
Determine z^n and all values of sqrts with base n of z.
z=(2-2i)^22 - (-1+isqrt(3))^33 , n=4
\(2-2i\\ = 2(1-i)\\ =2\sqrt2\left(\frac{1}{\sqrt2}+\frac{-i}{\sqrt2}\right)\\ =2\sqrt2\left(cos(\frac{-\pi}{4}+isin\frac{-\pi}{4}) \right)\\ =2\sqrt2e^{i*\frac{-\pi}{4}}\\ =2^{3/2}e^{-\pi/4}\)
\((2-2i)^{22}=\left[ 2^{3/2}e^{-(\pi /4)i} \right]^{22}\\ = 2^{33}e^{-(22\pi /4)i}\\ = 2^{33}e^{-(11\pi /2)i}\\ = 2^{33}e^{(-11\pi /2+12\pi/2)i}\qquad \mbox{I added 3 revolutions}\\ = 2^{33}e^{(\pi/2)i}\\ =2^{33}*i\)
\(- (-1+i\sqrt3)^{33}\\ =-(-1)^{33}(1-i\sqrt3)^{33}\\ =+(1-i\sqrt3)^{33}\\ =(\sqrt4)^{33}\left(\frac{1-i\sqrt3}{\sqrt{4}}{}\right)^{33}\\ =2^{33}\left(\frac{1-i\sqrt3}{2}{}\right)^{33}\\ =2^{33} \left[cos(-\pi/3)+isin((-\pi/3)i)\right]^{33}\\ =2^{33}\left[ e^{-\pi/3*i} \right]^{33}\\ =2^{33} e^{-11\pi *i} \\ =2^{33} e^{-11\pi *i+12\pi i} \\ =2^{33} e^{\pi i} \\ \)
\(=2^{33}*-1\\ =\;-2^{33}\)
\(z=2^{33}\;i-2^{33}\\ z=2^{33}\;(i-1)\\ ...\\ z^n=2^{33n}\;(i-1)^n\\ ....\\ z^4=2^{33*4}\;(i-1)^4\\ z^4=2^{132}\;(-1-2i+1)^2\\ z^4=2^{132}\;(-2i)^2\\ z^4=2^{132}\;*(-4)\\ z^4=\;-2^{132}\;*2^2\\ z^4=\;-2^{134}\\ z^4\approx \;-2.1778\times 10^{40} \)
\(\)
\(\)
(I did the second part first)
Why is this equality valid
2-2i = 2*sqrt(2)*e^-i*pi/4 = 2^3/2*e^-i*pi/4 ?
I will add some brackets here. You should have added the brackets yourself!
2-2i = 2*sqrt(2)*e^(-i*pi/4) = 2^(3/2)*e^(-i*pi/4) ? I assume that is what you want?
Firstly
\(2^{3/2}=2^{1+1/2}=2^1*2^{1/2}=2\sqrt2\)
so 2*sqrt(2)*e^(-i*pi/4) = 2^3/2*e^(-i*pi/4)
Now to show that they are the same as the first part.
i.e. show that
2-2i = 2*sqrt(2)*e^(-i*pi/4)
\(RHS=2*\sqrt2*e^{(-i*\pi /4)}\\ RHS=2*\sqrt2*e^{(-i*\pi /4)}\\ RHS=2*\sqrt2*[cos(-\pi /4)+isin(-\pi /4)]\\ RHS=2*\sqrt2*\left[\frac{1}{\sqrt2}+i*\frac{-1}{\sqrt2}\right]\\ RHS=2*\sqrt2*\left[\frac{1}{\sqrt2}-\frac{i}{\sqrt2}\right]\\ RHS=2(1-i)\\ RHS=2-2i\\ RHS=LHS \)
