z_k= e^i(2kπ/10)? z_k� z_j=e^{i(2π/10)(k+j)}, z_k is 10^th root of unity.
?\(\overline{z_k}\) is also 10^th root of unity. Taking z_j as \(\overline{z_k}\), we have z_k�z_j=1.

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).

The region formed is an annulus of inner radius 11 units and outer radius 19 units. Therefore, required area=π(19²?11²)=240π.

We can write: i=e^iπ/2?i^1/2=e^iπ/4=1/√2+i/?2, also i^1/2=e^i3π/4=-1/√2-i/√2
Hence, Re(i^1/2)+|Im(i^1/2)|=1/√2+1/√2=√2 OR -1/√2+1/√2=0.

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.

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).

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,
\(log?i=\frac{i\pi}{2}\)
\(i^i=e^{i(\frac{i\pi}{2})}=e^{\frac{-\pi}{2}}\).

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.