\(-\sqrt { 3 } +i\)

Modulus = 2 and
\(a={ tan }^{ -1 }\left| \frac { y }{ x } \right| ={ tan }^{ -1 }\frac { 1 }{ \sqrt { 3 } } =\frac { \pi }{ 6 } \)
Since the complex number  \(-\sqrt { 3 } +i\) lies in the second quadrant has the principal value
\(\theta =\pi -\alpha =\pi -\frac { \pi }{ 6 } =\frac { 5\pi }{ 6 } \)
Therefore the modulus and principal argument of \(-\sqrt { 3 } +i\) are 2 and \(\frac { 5\pi }{ 6 } \) respectively.

\(\left| \left( \overline { 1+i } \right) \left( 2+3i \right) \left( 4i-3 \right) \right| =\left| \left( \overline { 1+i } \right) \right| \left| 2+3i \right| \left| 4i-3 \right| \) (\(\because \) |z₁z₂z₃|=|z₁|z₂||z₃|)
= |1+i| |2+3i| |-3+4i|  \(\left( \because |z|=\left| \overline { z } \right| \right) \)
= \(\left( \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } \right) \left( \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 } } \right) \left( \sqrt { \left( 3 \right) ^{ 2 }+{ 4 }^{ 2 } } \right) \).
\(=(\sqrt{2})(\sqrt{13})(\sqrt{25})=5 \sqrt{26}\)

Let z = -6+8i
|z| =\(\sqrt { (-6)^{ 2 }+8^{ 2 } } \)
= \(\sqrt { 36+64 } =\sqrt { 100 } \) = 10
\(\sqrt { a+ib } =\pm \left( \sqrt { \frac { |z|+a }{ 2 } } +i\frac { b }{ |b| } \sqrt { \frac { |z|-a }{ 2 } } \right) \)
[Here |z| = 10, a = -6, b = 8]
\(\sqrt { -6+8i } \pm \left( \sqrt { \frac { 10-6 }{ 2 } } +i\frac { 8 }{ |8| } \sqrt { \frac { 10+6 }{ 2 } } \right) \)
= \(\pm \left( \sqrt { \frac { 4 }{ 2 } } +i\sqrt { \frac { 16 }{ 2 } } \right) \)
= \(\pm (\sqrt { 2 } +i\sqrt { 8 } )\)
= \(\\ \pm (\sqrt { 2 } +i2\sqrt { 2 } )\)
Aliter :
Square root of -6 + 8i
Let a + ib = - 6 + 8i
a = -6, b = 8
\(|z|=\sqrt{6^{2}+8^{2}}=\sqrt{100}=10\)
\(\sqrt{a+i b}=\pm\left[\sqrt{\frac{\sqrt{a^{2}+b^{2}}+a}{2}}+i \frac{b}{|b|} \sqrt{\frac{\sqrt{a^{2}+b^{2}}-a}{2}}\right]\)
\(=\pm\left[\sqrt{\frac{10-6}{2}}+i \sqrt{\frac{10+6}{2}}\right]\)
\(=\pm[\sqrt{2}+i \quad 2 \sqrt{2}]\)

2z+3w
= 2(5-2i)+3(-1+3i)
= 10-4i-3+9i
= (10-3)+ i(-4+9)
= 7+5i

i i² i³ ....i²⁰⁰⁰
= i^{1+2+3+.....+2000}
= \({ i }^{ \frac { 2000\times 2001 }{ 2 } }\)
[∴ 1+2+3+....n = \(\frac { n(n+1) }{ 2 } \)]
= i^{1000 x 2001}
= i^2001000
= 1
[∴ 2001000 is divisible by 4 as its last two digits are divisible by 4]