Given that z = 3 + 2i.
Therefore, iz = i(3 + 2i) = −2 + 3i
z + iz = (3 + 2i) + i(3 + 2i) = 1 + 5i
Let A,B, and C be z, z + iz, and iz respectively
\({ AB }^{ 2 }={ \left| (z+iz)-z \right| }^{ 2 }={ \left| -2+3i \right| }^{ 2 }=13\)
\({ BC }^{ 2 }{ =\left| iz-(z+iz) \right| }^{ 2 }={ \left| -3-2i \right| }^{ 2 }=13\)
\({ CA }^{ 2 }={ \left| z-iz \right| }^{ 2 }={ \left| 5-i \right| }^{ 2 }=26\)
Since AB² + BC² = CA² and AB = BC ΔABC is an isosceles right triangle.

We have |z| - |z - i|
\(\Rightarrow\)|x + iy| = |x + iy - i|
\(\Rightarrow\) \(\sqrt { { x }^{ 2 }+{ y }^{ 2 } } =\sqrt { { x }^{ 2 }+\left( y-1 \right) ^{ 2 } } \)
\(\Rightarrow x^{2}+y^{2}=x^{2}+y^{2}-2 y+1\)
\(\Rightarrow\) 2y -1 = 0

\(\left( cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right) \left( cos\frac { \pi }{ 12 } +isin\frac { \pi }{ 12 } \right) \)
[∵ (cosθ₁+isinθ₁)(cosθ₂+isonθ₂)
= cos(θ₁+θ₂)+isin(θ₁+θ₂)
= \(cos\left( \frac { 2\pi +\pi }{ 12 } \right) +isin\left( \frac { 2\pi +\pi }{ 12 } \right) \)
\(cos\left( \frac { 3\pi }{ 12 } \right) +isin\left( \frac { 3\pi }{ 12 } \right) \)
\(cos\left( \frac { \pi }{ 4 } \right) +isin\left( \frac { \pi }{ 4 } \right) \)
\(\frac { 1 }{ \sqrt { 2 } } +i\frac { 1 }{ \sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } \)(1+i)
Aliter:
\( \left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)\left(\cos \frac{\pi}{12}+i \sin \frac{\pi}{12}\right) \)
\( =\cos \left(\frac{\pi}{6}+\frac{\pi}{12}\right)+i \sin \left(\frac{\pi}{6}+\frac{\pi}{12}\right) \)
\( =\left(\cos \frac{3 \pi}{12}+i \sin \frac{3 \pi}{12}\right) \)
\( =\cos \frac{\pi}{4}+i \sin \frac{\pi}{4} \)
\( =\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}=\frac{1+i}{\sqrt{2}} \)

\(\left| z+3+4i \right| \le \left| z \right| +\left| 3+4i \right| =2+5=7\)
\(\left| z+3+4i \right| \le 7\) .............. (1)
\(\left| z+3+4i \right| \ge \left| \left| z \right| -\left| 3+4i \right| \right| =\left| 2-5 \right| =3\)
\(\left| z+3+4i \right| \ge 3\) ............ (2)
From (1) and (2) we get, \(3\le \left| z+3+4i \right| \le 7\) 

Let z₁ = (2+i)x + (1−i)y + 2i−3 = (2x+y−3) + i(x−y+ 2)and
z₂ = x+(−1+2i)y+1+i = (x−y+1) + i(2y+1)
Given that z₁ = z₂ 
Therefore (2x+y−3) + i(x−y+2) = (x−y+1) + i(2y+1).
Equating real and imaginary parts separately, gives
2x+y−3 = x−y+1 \(\Rightarrow\) x+2y = 4
x−y+2 = 2y +1 \(\Rightarrow\) x−3y = −1
Solving the above equations, gives
x = 2 and y = 1.

Given (x₁ + iy₁)(x₂ + iy₂) ....(x_n + iy_n) = a + ib
Taking modulus
|(x₁+ iy₁)(x₂ + iy₂) ....(x_n + iy_n)| = |a + ib|
|x₁ + iy₁| + |x₂ + iy₂|+...+ |x_n + iy_n| = |a + ib|
\( \sqrt{x_{1}^{2}+y_{1}^{2}} \sqrt{x_{2}^{2}+y_{2}^{2}} \sqrt{x_{3}^{2}+y_{3}^{2}} \cdots \sqrt{x_{n}{ }^{2}+y_{n}^{2}} =\sqrt{a^{2}+b^{2}} \)
Squaring on both sides
\( \left(x_{1}^{2}+\mathrm{y}_{1}^{2}\right)\left(x_{2}^{2}+\mathrm{y}_{2}^{2}\right)\left(x_{3}^{2}+\mathrm{y}_{3}^{2}\right) \ldots\left(x_{\mathrm{n}}^{2}+\mathrm{y}_{\mathrm{n}}^{2}\right) =\mathrm{a}^{2}+\mathrm{b}^{2} \)
Hence proved.

Since,\(\left| { z }_{ 1 } \right| =\left| { z }_{ 2 } \right| =\left| { z }_{ 3 } \right| =1\)
\(\left| { z }_{ 1 } \right| ^{ 2 }=1\Rightarrow { z }\bar { { z }_{ 1 } } =1,\left| { z }_{ 2 } \right| ^{ 2 }=1\Rightarrow { z }_{ 2 }\bar { { z }_{ 2 } } =1\ \)
Therefore, \(\bar { { z }_{ 1 } } =\frac { 1 }{ { z }_{ 1 } } ,\bar { { z }_{ 2 } } =\frac { 1 }{ { z }_{ 3 } } \) and hence
\(\left| \frac { 1 }{ { z }_{ 1 } } +\frac { 1 }{ { z }_{ 2 } } +\frac { 1 }{ { z }_{ 3 } } \right| =\left| \bar { { z }_{ 1 } } +\bar { { z }_{ 2 } } +{ \bar { z } }_{ 3 } \right| \)
= \(\left| \overline { { z }_{ 1 }+\left| { z }_{ 2 }+{ z }_{ 3 } \right| } \right| ={ { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } }=1\)

 i¹⁷²⁹ = i¹⁷²⁸ i¹ = i

\(\theta =\frac { 2\pi }{ 3 } \)
When θ = \(\frac { \pi }{ 3 } \)
Roration of z is \(ze^{ i\theta }=ze^{ i2\frac { \pi }{ 3 } }\) [using (1)]
= \(2\sqrt { 2 } e^{ -i\frac { \pi }{ 4 } }.{ e }^{ i2\frac { \pi }{ 3 } }\)   [using (1)]
= \(2\sqrt { 2 } e^{ i\left( i\frac { \pi }{ 3 } -\frac { \pi }{ 4 } \right) }=2\sqrt { 2 } e^{ i5\frac { \pi }{ 12 } }\)

\(\frac { 3 }{ 2 } \left( cos\frac { \pi }{ 3 } +isin\frac { \pi }{ 6 } \right) .6\left( cos\frac { 5\pi }{ 6 } +\frac { 5\pi }{ 6 } \right) \)
= \(\left( \frac { 3 }{ 2 } \right) \left( 6 \right) \left( cos\left( \frac { \pi }{ 3 } +\frac { 5\pi }{ 6 } \right) +isin\left( \frac { \pi }{ 3 } +\frac { 5\pi }{ 6 } \right) \right) \)
= \(9\left( cos\left( \frac { 7\pi }{ 6 } \right) +isin\left( \frac { 7\pi }{ 6 } \right) \right) \)
= \(9\left( cos\left( \pi +\frac { \pi }{ 6 } \right) +isin\left( \pi +\frac { \pi }{ 6 } \right) \right) \)
= \(9\left( -cos\left( \frac { \pi }{ 6 } \right) -isin\left( \frac { \pi }{ 6 } \right) \right) \)
= \(9\left( -\frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } \right) =\frac { 9\sqrt { 3 } }{ 2 } -\frac { 9i }{ 2 } \)