z=2i(x+iy)/(1-(x+iy)²)=2i(x+iy)/(1-(x²-y²+2ixy))
Using 1-x²=y², z=(2ix-2y)/(2y²-2ixy)=-1/y
? ?1≤y≤1 ⇒ ?1/y≤-1 or -1/y≥1.

f(x)=(x+5), if x>-5 and f(x)=-(x+5), if x≤-5
Since break is at x=-5, we calculate limit at this point.
We know that, a function is not differentiable at point x=a, if eir \(\lim_{h \to 0}\frac{f(a+h)-f( a )}{h}\)does not exist or is infinity. We check limits for both cases of function.
Left limit: Here, a=-5.
\(\lim_{h \to 0}\frac{f(-5+h)-f(-5)}{h}\)
\(\lim_{h \to 0}\frac{-(-5+h+5)-(-(-5+5))}{h}\)
=-1
Right limit: Here, a=1.
\(\lim_{h \to 0}\frac{f(1+h)-f(1)}{h}\)
\(\lim_{h \to 0}\frac{(1+h+5)-(1+5)}{h}\)
=1
Since two limits are not equal, function is not differentiable.

\(\frac{1}{x^2+x-132}\)=\(\frac{1}{(x-11)(x+12)}\) is discontinuous at x=11 since denominator becomes 0.
\(\frac{1}{x^3-7x^2-40x-44}\)=\(\frac{1}{(x-11)(x+2)^2}\)is discontinuous at x=11 since denominator becomes 0.
\(\frac{x-11}{x^3-23x^2+143x-121}\)=\(\frac{(x-11)}{(x-11)^2(x-1)}\)=\(\frac{1}{(x-11)(x-1)}\)is discontinuous at x=11 since denominator becomes 0.
\(\frac{1}{x^2-2x-143}\)is continuous at x=11.

Let y=1/(1-z) ⇒ z=1-1/y
|z|=1 ⇒ |1-1/y|=1 ⇒ |y-1|=|y| ⇒ locus of y is perpendicular bisector of line segment joining 0 and 1 ⇒ arg y ∈(-π/2, π/2).

For real x(≠0), let ix be root⇒x⁴-a₁x³i- a₂x²+a₃xi+a₄=0⇒x⁴-a₂x²+a₄=0 and a₁x³-a₃x=0
a₁x³-a₃x=0 ⇒ a₁x²-a₃=0 ⇒x²=a₃/a₁, putting this value in equation, a₃/(a₁a₂)+a₁a₄/(a₂a₃)=1.

Note that α=ω. Therefore, f(α)=f(ω)=(ω²?ω?1)⁷
=(ω²+ω²)⁷=(2ω²)⁷=2⁷ω¹⁴=128ω²=128α².

z=x+iy
In general limits are discussed at origin, if nothing is specified.
f(x,y)=(x+iy)²
Left limit: \(\lim_{x \to 0 \\ y \to 0}\)(x+iy)²
\(\lim_{y \to 0}\)i²y²
=0
Right limit: \(\lim_{y \to 0 \\ x \to 0}\)(x+iy)²
\(\lim_{x \to 0}\)x²
=0
Both limits are equal, refore function is continuous. To check differentiability,
f?(z)=\(\lim_{\delta z \to 0}\frac{f(z+\delta z)-f(z)}{\delta z}\)should exist.
\(\lim_{\delta z \to 0}\frac{(z+\delta z)^2-(z)^2}{\delta z}\)
\(\lim_{\delta z \to 0}\frac{2z(\delta z)+(\delta z)^2}{\delta z}\)
=\(\lim_{\delta z \to 0}\)(2z+(δz))
=2z
Since f?(z) exists, function is differentiable as well.

When we substitute x=0, x=5 or x=-2 in f(x), we get denominator as 0. This is because se are points of discontinuity.
At x=2, we get f(2)=\(\frac{-1}{24}\) which is continuous.

f(z)=az²+b, with a, b of same sign ⇒ f(f(z))=a(az²+b)²+b
If z∈R or iz∈R ⇒ z²∈R ⇒ f(z)∈R ⇒ f(f(z))≠0 ⇒ Hence real or purely imaginary number cannot satisfy f(f(z))=0.

\(\frac{x^2-81}{x-9}\)=\(\frac{(x-9)(x+9)}{x-9}\)=(x-9) which is continuous at x = 9.
\(\frac{x^3-20x^2+117x-162}{x-9}\)=\(\frac{(x-9)^2(x-2)}{(x-9)}\)=(x-9)(x-2) which is continuous at x = 9.
\(\frac{x^2-11x+18}{x(x-9)}\)=\(\frac{(x-2)(x-9)}{x(x-9)}\)=\(\frac{(x-2)}{x}\) which is continuous at x = 9.
\(\frac{x^2-13x-90}{x-9}\)=\(\frac{(x-18)(x+5)}{x-9}\) which is discontinuous at x = 9, since denominator becomes 0.