Answer to Question #106692 in Calculus for Parul

Question #106692
a) Obtain the 6th roots of (-7) , represent them in an Argand diagram.
b) using epsilon Delta definition , show that Lim(3x-5)=1
x->2
1
Expert's answer
2020-04-01T14:14:32-0400

a) The 6th root of "(-7)" are the solution of the equation

"z^6+7=0" ............(1)

Where z is a complex number.


"z^6+7=0 \\ \\implies z^6=-7=7(-1) \\ \\implies z=7^{\\frac{1}{6}}(-1)^{\\frac{1}{6}}"

Again ,we known that

"e^{\\pi i}=cos(\\pi)+i sin(\\pi)"

"=-1"

Again,"e^{2\\pi k i}=1" for all "k\\in \\Z"

Therefore,

"e^{\\pi i}e^{2k\\pi i}=-1 \\ \\implies e^{(2k+1)\\pi i }=-1"

for all "k\\in \\Z" .


"\\therefore z= 7^{\\frac{1}{6}} e^{\\frac{(2k+1)\\pi i }{6}}"

"Where \\ , k=0,1,2,3,4,5" give the solution of equation (1).

Therefore ,the 6th root of "(-7)" are


"\\alpha e^{\\frac{\\pi i} {6}},\\alpha e^{\\frac{3\\pi i}{6}},\\alpha e^{\\frac{5\\pi i}{6}},\n\\alpha e^{\\frac{7\\pi i}{6}},\\alpha e^{\\frac{9\\pi i}{6}},\\alpha e^{\\frac{11\\pi i}{6}}"


Where "\\alpha=7^{\\frac{1}{6}}" .

The argand diagram is at following photo.






b) Let "f(x)=3x-5 \\ ,\\forall \\ x\\in \\R"

A little algebraic manipulation gives us

"\\mid f(x)-1 \\mid=\\mid 3x-5-1\\mid=\\mid 3(x-2)\\mid=3\\mid x-2\\mid."

Now , for a given "\\epsilon \\gt 0" ,we choose "\\delta=\\frac{\\epsilon}{3}" . Then if "\\mid x-2\\mid \\lt\\delta,"

"\\mid f(x)-1\\mid=\\mid3x-5-1\\mid= \\mid 3x-6\\mid=3\\mid x-2\\mid \\lt3\\delta=\\epsilon"

Since "\\epsilon \\gt0" is arbitrary, we deduce that "\\lim_{x \\to 2}f(x)=1"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS