i²i³...i⁴⁰ = i^1+2+3...+40 = \(i^{\frac{40 \times 41}{2}}=i^{820}=i^{0}=1 .\)

 Im(3z+4\(\bar { z } \)−4i)
= Im (3(x+iy)+4(x-iy)-4i)
= Im (3x+i3y+4x-i4y-4i)
= Im (3x+4x+i(3y-4y-4)
= Im (7x+i(-y-4))
∴ Imaginary part is -y - 4

\({ i }^{ 59 }+\frac { 1 }{ { i }^{ 59 } } \)
i^{4 \(\times\) 14 + 3} + i^{-(4 \(\times\) 14 + 3)}
= (i⁴)¹⁴.i³ + (i⁴)^-14.i^-3
= 1.i³+1.i^-3  [∵ i⁴ = 1]
= -i + i  [∴ i³ = -i and i^-3= i]
= 0

\(\left| z-2-i \right| =3\)
⇒ |z-(2+i)| = 3
It is of the form |z - z₀| = r and so it represents a circle.
Centre is (2, 1) and radius = 3 units.
Aliter:
Let z = x +iy
|z-2-i| = 3
|x + iy-2-i| = 3
\(|(x-2)+i(y-1)|=3\)
\(\sqrt{(x-2)^{2}+(y-1)^{2}}=3\)
Squaring on both sides
\( (x-2)^{2}+(y-1)^{2}=9 \)
\(x^{2}-4 x+4+y^{2}-2 y+1-9=0 \)
\(x^{2}+y^{2}-4 x-2 y-4=0 \)
Comparing with General form of equation of circle
\(a x^{2}+b y^{2}+2 g x+2 f y+c=0\)
we get a = 1, b = 1, g = -2, f = -1, c = -4
Centre (- g, - f) = (2, 1)
radius = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{4+1+4}=\sqrt{9}\)
= 3 units

2 +i2\(\sqrt { 3 } \)
Let 2+i2\(\sqrt { 3 } \)  = x + iy = r (cosθ + i sinθ)
r = modulus =\(\\ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } \)
=\(\\ \sqrt { { 2 }^{ 2 }+(2\sqrt { 3 } )^{ 2 } } \)
= \(\sqrt { 4+12 } =\sqrt { 16 } \) = 4
α = tan-1\(\left| \frac { y }{ x } \right| =tan^{ -1 }\left| \frac { 2\sqrt { 3 } }{ 2 } \right| \)
= \(tan^{ -1 }(\sqrt { 3 } )=\frac { \pi }{ 3 } \)
Since the complex number 2+i2 \(\sqrt { 3 } \) lies in the I quadrant, [x, y both +ve] its principal value θ = α = \(\frac { \pi }{ 3 } \)
∴ Its polar form is 2+i2\(\sqrt { 3 } \)
= 4\(\left[ cos\left( 2k\pi +\frac { \pi }{ 3 } \right) +isin\left( 2k\pi +\frac { \pi }{ 3 } \right) \right] ,k\in Z\).