Answer to Question #106692 in Calculus for Parul

Question

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

Answer 1

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

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

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 $\lim_{x \to 2}f(x)=1$