Since ω=z̅+4i and z=6+8i (from image), refore ω=6-8i+4i=6-4i and we get a triangle having vertices (0,0), (6,8) and (6,-4) in Argand Plane shown. Therefore, area is (1/2)�(base)�(height)=(1/2)�(8+4)�6=36.

lnz+lnω=ln|z|+iargz+ln|ω|+iargω
Since, |z|=|ω| and arg z+arg ω=0, ∴ lnz+lnω=2ln|z|
∴ Required value=2ln|z|/ln|z|=2.

If we draw a circle with z₂-z₁ as a diameter, we see that z₃ would be outside circle, since radius of circle is 5(<6). But we know that points on circle only would result in an angle of π/2. Hence, no such z₃ is possible.

We know that
\(a^x=e^{x loga}\)
\(i^i=e^{i log?i}\)
We also know from definition of logarithm,
\(log?i=\frac{i\pi}{2}\)
\(i^i=e^{i(\frac{i\pi}{2})}=e^{\frac{-\pi}{2}}\).

Given relations can be written as: 1/(a+ω)+1/(b+ω)+1/(c+ω)=2/ω, and
1/(a+ω²)+1/(b+ω²)+1/(c+ω²)=2/ω²?ω and ω² are roots of 1/(a+x)+1/(b+x)+1/(c+x)=2/x.
⇒[3x²+2(a+b+c)x+bc+ca+ab]/[(a+x)(b+x)(c+x)]=2/x⇒x³+(bc+ca+ab)x-2abc=0.
Let 3^rd root be α (apart from ω and ω²). Then, α+ ω+ ω²=coefficient of x²=0⇒α=1. Hence, 1/(a+1)+1/(b+1)+1/(c+1)=2/1=2.

|z-1|<|z-3| ⇒ (x-1)²+y²<(x-3)²+y² ⇒ x<2.
Therefore, a triangle is formed with base length 4 and height 2 (along x-axis). Hence, required area=1/2�4�2=4.

We know that,
\(1-i=\sqrt 2 (\frac{1}{\sqrt 2}-\frac{i}{\sqrt 2})=\sqrt 2 (cos \frac{\pi}{4}-isin \frac{\pi}{4})\)
\((1-i) ^{100}=(\sqrt 2 (cos \frac{\pi}{4}-isin \frac{\pi}{4})) ^{100}=2^{50}((cos \frac{\pi}{4}-isin \frac{\pi}{4})) ^{100}\)
By Applying DeMoivre?s Theorem
\((1-i)^{100}=2^{50} (cos \frac{100\pi}{4}-isin \frac{100\pi}{4})\)
\((1-i)^{100}=2^{50} (cos?25\pi-sin25\pi)\).

(a+b)+(2a+b)ω+(3a+b)ω²+?+(na+b)ω^n-1=b(1+ω+ω²+….+ω^n-1)+a(1+2ω+3ω²+?+nω^n-1)
Let S=1+2ω+3ω²+?+nω^n-1?Sω= ω+2ω²+?+(n-1)ω^n-1+nωⁿ⇒S(1-ω)=1+ω+ ω²+?+ω^n-1?nωⁿ=-n
⇒S=n/(ω-1)⇒E=na/(ω-1).

We have, x²-3x+2=x²-6x+8 and y²-11y+40+y²-9x+10=0
Hence, x=2, y=5
∴ |2+5i|=29^1/2.

if x=y or x=-y, |Arg(z)|=π/4 and α=2� π/4=π/2.
Now, if |y|>|x|, n |Arg(z)|>π/4 and α>2�π/4=π/2.
If z=i-1, n, Arg(z)=3π/4 but α=π/2, ∴ α≠2�Arg(z).