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Complex plane

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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 ?

Guest Nov 18, 2015
edited by Guest  Nov 18, 2015
edited by Guest  Nov 18, 2015

#3
+91038
+5

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)

$$RHS=2*\sqrt2*e^{(-i*\pi /4)}\\ RHS=2*\sqrt2*e^{(-i*\pi /4)}\\ RHS=2*\sqrt2*[cos(-\pi /4)+isin(-\pi /4)]\\ RHS=2*\sqrt2*\left[\frac{1}{\sqrt2}+i*\frac{-1}{\sqrt2}\right]\\ RHS=2*\sqrt2*\left[\frac{1}{\sqrt2}-\frac{i}{\sqrt2}\right]\\ RHS=2(1-i)\\ RHS=2-2i\\ RHS=LHS$$

Melody  Nov 18, 2015
edited by Melody  Nov 19, 2015
edited by Melody  Nov 19, 2015
edited by Melody  Nov 19, 2015
edited by Melody  Nov 19, 2015
edited by Melody  Nov 19, 2015
edited by Melody  Nov 19, 2015
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#1
0

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!.

Guest Nov 18, 2015
#2
+5

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

$$\cos(-\pi/4)=1/\sqrt{2}$$,

$$\sin(-\pi/4)=-1/\sqrt{2}.$$

Guest Nov 18, 2015
#3
+91038
+5

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)

$$RHS=2*\sqrt2*e^{(-i*\pi /4)}\\ RHS=2*\sqrt2*e^{(-i*\pi /4)}\\ RHS=2*\sqrt2*[cos(-\pi /4)+isin(-\pi /4)]\\ RHS=2*\sqrt2*\left[\frac{1}{\sqrt2}+i*\frac{-1}{\sqrt2}\right]\\ RHS=2*\sqrt2*\left[\frac{1}{\sqrt2}-\frac{i}{\sqrt2}\right]\\ RHS=2(1-i)\\ RHS=2-2i\\ RHS=LHS$$

Melody  Nov 18, 2015
edited by Melody  Nov 19, 2015
edited by Melody  Nov 19, 2015
edited by Melody  Nov 19, 2015
edited by Melody  Nov 19, 2015
edited by Melody  Nov 19, 2015
edited by Melody  Nov 19, 2015
#5
+91038
0

Finallly I am happy with my answer.

Melody  Nov 19, 2015

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