1. $\sqrt{-3}$

   Solution: $\sqrt{-3}$

   Since $\sqrt{-1}=i$, we have:

   $\Rightarrow \sqrt{-3}=\sqrt{3\cdot(-1)}=\sqrt{3}\cdot\sqrt{-1}=\sqrt{3}i$; [$\because\ i=\sqrt{-1}$]

2. $6\sqrt{-4}$

   Solution: $6\sqrt{-4}$

   $\Rightarrow 6\sqrt{-4}=6\sqrt{4\cdot(-1)}=6\cdot\sqrt{4}\cdot\sqrt{-1}=6\cdot2\cdot i=12i$; [$\because\ i=\sqrt{-1}$]

3. $\sqrt{-\frac{4}{9}}$

   Solution: $\sqrt{-\frac{4}{9}}$

   $\Rightarrow \sqrt{-\frac{4}{9}}=\sqrt{\frac{4}{9}\cdot(-1)}=\sqrt{\frac{4}{9}}\cdot\sqrt{-1}=\frac{2}{3}i$; [$\because\ i=\sqrt{-1}$]

4. $-\sqrt{-20}$

   Solution: $-\sqrt{-20}$

   $\Rightarrow -\sqrt{-20}=-\sqrt{20\cdot(-1)}=-\sqrt{20}\cdot\sqrt{-1}$

   $=\sqrt{4}\cdot\sqrt{5}\cdot\sqrt{-1}=2\sqrt{5}\cdot\sqrt{-1}=-2\sqrt{5}i$

5. $4-\sqrt{-60}$

   Solution: $4-\sqrt{-60}$

   $\Rightarrow 4-\sqrt{-60}=4-\sqrt{60\cdot(-1)}=4-\sqrt{60}\cdot\sqrt{-1}=4-\sqrt{4}\cdot\sqrt{15}\cdot\sqrt{-1}$

   $=4-\sqrt{4}\cdot\sqrt{15}\cdot\sqrt{-1}=4-2\sqrt{15}i$

6. $\sqrt{-8}\,\sqrt{-2}$

   Solution: $\sqrt{-8}\,\sqrt{-2}$

   $\Rightarrow \sqrt{-8}\,\sqrt{-2}=(\sqrt{8}\cdot\sqrt{-1})(\sqrt{2}\cdot\sqrt{-1})=(\sqrt{8}i)(\sqrt{2}i)$

   $=\sqrt{8}\cdot\sqrt{2}\cdot i^2=\sqrt{16}\cdot i^2=4\cdot(-1)=-4$; [$\because\ i^2=-1$]

1. $(4-i)+(5+5i)$

   Solution: $(4-i)+(5+5i)$

   We can combine the real parts and the imaginary parts separately:

   $\Rightarrow (4-i)+(5+5i)=(4+5)+(-i+5i)=9+4i$

2. $(7-6i)-(5-6i)$

   Solution: $(7-6i)-(5-6i)$

   We can combine the real parts and the imaginary parts separately:

   $\Rightarrow (7-6i)-(5-6i)=7-6i-5+6i$

   Therefore, the simplified expression is $2+0i=2$.

3. $(-2+8i)-(7+3i)$

   Solution: $(-2+8i)-(7+3i)$

   We can combine the real parts and the imaginary parts separately:

   $\Rightarrow (-2+8i)-(7+3i)=-2+8i-7-3i=-9+5i$

4. $(4-2i)-(5-2i)$

   Solution: $(4-2i)-(5-2i)$

   We can combine the real parts and the imaginary parts separately:

   $\Rightarrow (4-2i)-(5-2i)=4-2i-5+2i=-1+0i=-1$

5. $(2+4i)(1+2i)$

   Solution: $(2+4i)(1+2i)$

   $\Rightarrow (2+4i)(1+2i)=2(1)+2(2i)+4i(1)+4i(2i)=2+4i+4i+8i^2$

   Since $i^2=-1$, we can substitute that into the expression:

   $\Rightarrow 2+4i+4i+8(-1)=2+8i-8$

   Now, combine the real and imaginary parts:

   $\Rightarrow (2-8)+8i=-6+8i$

6. $(1-4i)(2-3i)$

   Solution: $(1-4i)(2-3i)$

   $\Rightarrow (1-4i)(2-3i)=1(2)+1(-3i)-4i(2)-4i(-3i)$

   $=2-3i-8i+12i^2$

   Since $i^2=-1$, we can substitute that into the expression:

   $\Rightarrow 2-3i-8i+12(-1)=2-11i-12$

   Now, combine the real and imaginary parts:

   $\Rightarrow (2-12)-11i=-10-11i$

7. $-8i(2-2i)$

   Solution: $-8i(2-2i)$

   $\Rightarrow -8i(2-2i)=-8i(2)-8i(-2i)=-16i+16i^2$

   Since $i^2=-1$, we can substitute that into the expression:

   $\Rightarrow -16i+16(-1)=-16i-16$

   Now, rewrite the expression in the standard form $a+bi$:

   $=-16-16i$

8. $(3+2i)^2$

   Solution: $(3+2i)^2$

   $\Rightarrow (3+2i)^2=(3+2i)(3+2i)$

   $\Rightarrow (3+2i)(3+2i)=3(3)+3(2i)+2i(3)+2i(2i)=9+6i+6i+4i^2$

   Since $i^2=-1$, we can substitute that into the expression:

   $\Rightarrow 9+6i+6i+4(-1)=9+12i-4$

   Now, combine the real and imaginary parts:

   $\Rightarrow (9-4)+12i=5+12i$

9. $(3-6i)(3+6i)$

   Solution: $(3-6i)(3+6i)$

   $\Rightarrow (3-6i)(3+6i)=3(3)+3(6i)-6i(3)-6i(6i)$

   $=9+18i-18i-36i^2$

   Since $i^2=-1$, we can substitute that into the expression:

   $\Rightarrow 9+18i-18i-36(-1)=9-36(-1)=9+36=45$

10. $(-5-3i)^2$

    Solution: $(-5-3i)^2$

    $\Rightarrow (-5-3i)^2=(-5-3i)(-5-3i)$

    $\Rightarrow (-5-3i)(-5-3i)=(-5)(-5)+(-5)(-3i)+(-3i)(-5)+(-3i)(-3i)$

    $=25+15i+15i+9i^2$

    Since $i^2=-1$, we can substitute that into the expression:

    $\Rightarrow 25+15i+15i+9(-1)=25+30i-9$

    Now, combine the real and imaginary parts:

    $\Rightarrow (25-9)+30i=16+30i$

11. $(1+\sqrt{2}i)(1-\sqrt{3}i)$

    Solution: $(1+\sqrt{2}i)(1-\sqrt{3}i)$

    $\Rightarrow (1+\sqrt{2}i)(1-\sqrt{3}i)=1(1)+1(-\sqrt{3}i)+\sqrt{2}i(1)+\sqrt{2}i(-\sqrt{3}i)$

    $=1-\sqrt{3}i+\sqrt{2}i-\sqrt{6}i^2$

    Since $i^2=-1$, we can substitute that into the expression:

    $\Rightarrow 1-\sqrt{3}i+\sqrt{2}i-\sqrt{6}(-1)=1-\sqrt{3}i+\sqrt{2}i+\sqrt{6}$

    Now, combine the real and imaginary parts:

    $\Rightarrow (1+\sqrt{6})+(-\sqrt{3}+\sqrt{2})i=(1+\sqrt{6})+(\sqrt{2}-\sqrt{3})i$

12. $(\sqrt{2}+i)(\sqrt{2}-i)$

    Solution: $(\sqrt{2}+i)(\sqrt{2}-i)$

    $\Rightarrow (\sqrt{2}+i)(\sqrt{2}-i)=(\sqrt{2})(\sqrt{2})+(\sqrt{2})(-i)+i(\sqrt{2})+i(-i)$

    $=2-\sqrt{2}i+\sqrt{2}i-i^2$

    Since $i^2=-1$, we can substitute that into the expression:

    $\Rightarrow 2-\sqrt{2}i+\sqrt{2}i-(-1)=2+1=3$

1. $i^9$

   Solution: $i^9$

   $\Rightarrow i^9=i^8\cdot i=(i^2)^4\cdot i^1$

   Since $i^2=-1$, we have:

   $\Rightarrow i^9=(-1)^4\cdot i=1\cdot i=i$

2. $i^{13}$

   Solution: $i^{13}$

   $\Rightarrow i^{13}=i^{12}\cdot i=(i^2)^6\cdot i^1$

   Since $i^2=-1$, we have:

   $\Rightarrow i^{13}=(-1)^6\cdot i=1\cdot i=i$

3. $i^{28}$

   Solution: $i^{28}$

   $\Rightarrow i^{28}=(i^2)^{14}$

   Since $i^2=-1$, we have:

   $\Rightarrow i^{28}=(-1)^{14}=1$

4. $(-i)^{21}$

   Solution: $(-i)^{21}$

   $\Rightarrow (-i)^{21}=(-1)^{21}\cdot(i)^{21}$

   Since $(-1)^{21}=-1$, we have:

   $\Rightarrow (-i)^{21}=-i^{21}$

   Now, we need to simplify $i^{21}$. We can express $i^{21}$ as follows:

   $\Rightarrow -i^{21}=-i^{20}\cdot i=-(i^2)^{10}\cdot i$

   Since $i^2=-1$, we have:

   $=-(-1)^{10}\cdot i=-1\cdot i=-i$

5. $(3i)^3$

   Solution: $(3i)^3$

   $=27\cdot i^3$

   Substitute the simplified values: Since $i^2=-1$, we have:

   $\Rightarrow 27\cdot i^3=27i^2\cdot i=27\cdot(-1)\cdot i=-27i$

6. $(-2i)^4$

   Solution: $(-2i)^4$

   $(-2i)^4=16\cdot i^4=16(i^2)^2=16(-1)^2=16$; [$\because\ i^2=-1$]

1. $9+i^6$

   Solution: $9+i^6$

   $\Rightarrow 9+i^6=9+(i^2)^3=9+(-1)^3=9-1=8$; [$\because\ i^2=-1$]

   Since there is no imaginary part, we can write this as $8+0i$.

2. $-17+i^5$

   Solution: $-17+i^5$

   $\Rightarrow -17+i^5=-17+i^4\cdot i=-17+(i^2)^2\cdot i$

   $=-17+(-1)^2\cdot i=-17+i$; [$\because\ i^2=-1$]

3. $-13i$

   Solution: $i^4-13i$

   $\Rightarrow i^4-13i=(i^2)^2-13i=(-1)^2-13i=1-13i$; [$\because\ i^2=-1$]

4. $i^5+21i$

   Solution: $i^5+21i$

   $\Rightarrow i^5+21i=i^4\cdot i+21i=(i^2)^2\cdot i+21i=(-1)^2\cdot i+21i$

   $=1\cdot i+21i=i+21i=22i$; [$\because\ i^2=-1$]

   Since there is no real part, we can write this as $0+22i$.

5. $i^5+i^7$

   Solution: $i^5+i^7$

   $\Rightarrow i^5+i^7=i^4\cdot i+i^6\cdot i=(i^2)^2\cdot i+(i^2)^3\cdot i$; [$\because\ i^2=-1$]

   $=(-1)^2\cdot i+(-1)^3\cdot i=i+(-i)=i-i=0$

   Since there is no imaginary part, we can write this as $0+0i$.

6. $i^{74}-i^{100}$

   Solution: $i^{74}-i^{100}$

   $\Rightarrow i^{74}-i^{100}=(i^2)^{37}-(i^2)^{50}=(-1)^{37}-(-1)^{50}$; [$\because\ i^2=-1$]

   $=-1-1=-2$

   Since there is no imaginary part, we can write this as $-2+0i$.

1. $\dfrac{3}{4-i}$

   Solution: $\dfrac{3}{4-i}$

   We need to multiply the numerator and denominator by the conjugate of the denominator. The conjugate of $4-i$ is $4+i$.

   $\Rightarrow \dfrac{3}{4-i}=\dfrac{3}{4-i}\times\dfrac{4+i}{4+i}=\dfrac{3(4+i)}{(4-i)(4+i)}=\dfrac{12+3i}{(4)^2-i^2}$

   Since $i^2=-1$, we can substitute this into the denominator:

   $\Rightarrow \dfrac{12+3i}{16-(-1)}=\dfrac{12+3i}{16+1}=\dfrac{12+3i}{17}$

   Finally, separate the real and imaginary parts:

   $\Rightarrow \dfrac{12+3i}{17}=\dfrac{12}{17}+\dfrac{3}{17}i$

2. $\dfrac{3i}{6+5i}$

   Solution: $\dfrac{3i}{6+5i}$

   We need to multiply the numerator and denominator by the conjugate of the denominator. The conjugate of $6+5i$ is $6-5i$.

   $\Rightarrow \dfrac{3i}{6+5i}=\dfrac{3i}{6+5i}\times\dfrac{6-5i}{6-5i}=\dfrac{3i(6-5i)}{(6+5i)(6-5i)}$

   $\Rightarrow \dfrac{3i(6-5i)}{(6+5i)(6-5i)}=\dfrac{18i-15i^2}{(6)^2-25i^2}$

   Since $i^2=-1$, we can substitute this into the numerator and denominator:

   $\Rightarrow \dfrac{18i-15(-1)}{36-25(-1)}=\dfrac{18i+15}{36+25}=\dfrac{15+18i}{61}$

   Finally, separate the real and imaginary parts:

   $\Rightarrow \dfrac{15+18i}{61}=\dfrac{15}{61}+\dfrac{18}{61}i$

3. $\dfrac{3-i\sqrt{5}}{3+i\sqrt{5}}$

   Solution: $\dfrac{3-i\sqrt{5}}{3+i\sqrt{5}}$

   We need to multiply the numerator and denominator by the conjugate of the denominator. The conjugate of $3+i\sqrt{5}$ is $3-i\sqrt{5}$.

   $\Rightarrow \dfrac{3-i\sqrt{5}}{3+i\sqrt{5}}=\dfrac{3-i\sqrt{5}}{3+i\sqrt{5}}\times\dfrac{3-i\sqrt{5}}{3-i\sqrt{5}}=\dfrac{(3-i\sqrt{5})^2}{(3+i\sqrt{5})(3-i\sqrt{5})}$

   Now, multiply out the numerator and denominator:

   $\Rightarrow \dfrac{(3-i\sqrt{5})^2}{(3+i\sqrt{5})(3-i\sqrt{5})}=\dfrac{(3)^2-2\times3\times i\sqrt{5}+(i\sqrt{5})^2}{(3)^2-(i\sqrt{5})^2}$

   $=\dfrac{9-6i\sqrt{5}+5i^2}{9-(-5i^2)}=\dfrac{9-6i\sqrt{5}-5}{9-(-5i^2)}=\dfrac{4-6i\sqrt{5}}{14}$; [$\because\ i^2=-1$]

   Finally, separate the real and imaginary parts:

   $\Rightarrow \dfrac{4-6i\sqrt{5}}{14}=\dfrac{4}{14}-\dfrac{6\sqrt{5}}{14}i=\dfrac{2}{7}-\dfrac{3\sqrt{5}}{7}i$

4. $\dfrac{2+7i}{5i}$

   Solution: $\dfrac{2+7i}{5i}$

   We need to multiply the numerator and denominator by the conjugate of the denominator. The conjugate of $5i$ is $-5i$.

   $\Rightarrow \dfrac{2+7i}{5i}=\dfrac{2+7i}{5i}\times\dfrac{-5i}{-5i}=\dfrac{(2+7i)(-5i)}{(5i)(-5i)}$

   Now, multiply out the numerator and denominator:

   $\Rightarrow \dfrac{(2+7i)(-5i)}{(5i)(-5i)}=\dfrac{-10i-35i^2}{-25i^2}$

   Since $i^2=-1$, we can substitute this into the numerator and denominator:

   $\Rightarrow \dfrac{-10i-35(-1)}{-25(-1)}=\dfrac{-10i+35}{25}=\dfrac{35-10i}{25}$

   Finally, separate the real and imaginary parts:

   $\Rightarrow \dfrac{35-10i}{25}=\dfrac{35}{25}-\dfrac{10}{25}i=\dfrac{7}{5}-\dfrac{2}{5}i$

5. $\dfrac{4+5i}{4-5i}$

   Solution: $\dfrac{4+5i}{4-5i}$

   We need to multiply the numerator and denominator by the conjugate of the denominator. The conjugate of $4-5i$ is $4+5i$.

   $\Rightarrow \dfrac{4+5i}{4-5i}=\dfrac{4+5i}{4-5i}\times\dfrac{4+5i}{4+5i}=\dfrac{(4+5i)^2}{(4-5i)(4+5i)}$

   Now, multiply out the numerator and denominator:

   $\Rightarrow \dfrac{(4+5i)^2}{(4-5i)(4+5i)}=\dfrac{(4)^2+40i+25i^2}{(4)^2-25i^2}$

   Since $i^2=-1$, we can substitute this into the numerator and denominator:

   $\Rightarrow \dfrac{16+40i+25(-1)}{16-25(-1)}=\dfrac{16+40i-25}{16+25}=\dfrac{-9+40i}{41}$

   Finally, separate the real and imaginary parts:

   $\Rightarrow \dfrac{-9+40i}{41}=-\dfrac{9}{41}+\dfrac{40}{41}i$

6. $\dfrac{3+2i}{2+i}$

   Solution: $\dfrac{3+2i}{2+i}$

   We need to multiply the numerator and denominator by the conjugate of the denominator. The conjugate of $2+i$ is $2-i$.

   $\Rightarrow \dfrac{3+2i}{2+i}=\dfrac{3+2i}{2+i}\times\dfrac{2-i}{2-i}=\dfrac{(3+2i)(2-i)}{(2+i)(2-i)}$

   Now, multiply out the numerator and denominator:

   $\Rightarrow \dfrac{(3+2i)(2-i)}{(2+i)(2-i)}=\dfrac{6-3i+4i-2i^2}{(2)^2-i^2}$

   Since $i^2=-1$, we can substitute this into the numerator and denominator:

   $\Rightarrow \dfrac{6-3i+4i-2(-1)}{4-(-1)}=\dfrac{6+i+2}{4+1}=\dfrac{8+i}{5}$

   Finally, separate the real and imaginary parts:

   $\Rightarrow \dfrac{8+i}{5}=\dfrac{8}{5}+\dfrac{1}{5}i$

7. $\dfrac{5+i}{1+2i}$

   Solution: $\dfrac{5+i}{1+2i}$

   We need to multiply the numerator and denominator by the conjugate of the denominator. The conjugate of $1+2i$ is $1-2i$.

   $\Rightarrow \dfrac{5+i}{1+2i}=\dfrac{5+i}{1+2i}\times\dfrac{1-2i}{1-2i}=\dfrac{(5+i)(1-2i)}{(1+2i)(1-2i)}$

   Now, multiply out the numerator and denominator:

   $\Rightarrow \dfrac{(5+i)(1-2i)}{(1+2i)(1-2i)}=\dfrac{5-10i+i-2i^2}{(1)^2-4i^2}$

   Since $i^2=-1$, we can substitute this into the numerator and denominator:

   $\Rightarrow \dfrac{5-9i-2(-1)}{1-4(-1)}=\dfrac{5-9i+2}{1+4}=\dfrac{7-9i}{5}$

   Finally, separate the real and imaginary parts:

   $\Rightarrow \dfrac{7-9i}{5}=\dfrac{7}{5}-\dfrac{9}{5}i$

8. $\dfrac{a+ib}{a-ib}$

   Solution: $\dfrac{a+ib}{a-ib}$

   We multiply the numerator and the denominator by the conjugate of the denominator, which is $a+ib$.

   $\Rightarrow \dfrac{a+ib}{a-ib}=\dfrac{(a+ib)(a+ib)}{(a-ib)(a+ib)}$

   $=\dfrac{a^2+2abi+(ib)^2}{a^2-(ib)^2}=\dfrac{a^2+2abi+i^2b^2}{a^2-i^2b^2}$

   Since $i^2=-1$, we have:

$\Rightarrow \dfrac{a^2+2abi-b^2}{a^2+b^2}=\dfrac{(a^2-b^2)+2abi}{a^2+b^2}$

Finally, separate the real and imaginary parts:

$=\dfrac{a^2-b^2}{a^2+b^2}+\dfrac{2ab}{a^2+b^2}i$

9. $\dfrac{1+i}{(1-i)^2}$

   Solution: $\dfrac{1+i}{(1-i)^2}\quad \ldots\ (1)$

   First, we expand the denominator:

   $\Rightarrow (1-i)^2=(1-i)(1-i)=1-i-i+i^2=1-2i+i^2$

   Since $i^2=-1$, we have:

   $\Rightarrow (1-i)^2=1-2i-1=-2i$

   So the expression (1) becomes: $\dfrac{1+i}{-2i}$

   Now, we multiply the numerator and the denominator by the conjugate of the denominator, which is:

   $\Rightarrow \dfrac{1+i}{-2i}=\dfrac{(1+i)(i)}{(-2i)(i)}=\dfrac{i+i^2}{-2i^2}\quad\Rightarrow\quad\dfrac{i-1}{-2(-1)}=\dfrac{i-1}{2}=\dfrac{-1+i}{2}\ ;\quad [\because\ i^2=-1]$

   Finally, separate the real and imaginary parts:

   $=-\dfrac{1}{2}+\dfrac{1}{2}i$

10. $\dfrac{(2+2i)^2}{(1+i)^2}$

    Solution: $\dfrac{(2+2i)^2}{(1+i)^2}\quad \ldots\ (1)$

    First, we expand the numerator and the denominator separately:

    $\Rightarrow (2+2i)^2=(2+2i)(2+2i)=4+4i+4i+4i^2=4+8i+4i^2$

    Since $i^2=-1$, we have:

    $\Rightarrow (2+2i)^2=4+8i-4=8i$

    $\Rightarrow (1+i)^2=(1+i)(1+i)=1+i+i+i^2=1+2i+i^2$

    Since $i^2=-1$, we have:

    $\Rightarrow (1+i)^2=1+2i-1=2i$

    So the expression (1) becomes: $\dfrac{8i}{2i}$

    Now, we can simplify the fraction by canceling the common factor $2i$.

    $\Rightarrow \dfrac{8i}{2i}=\dfrac{8}{2}=4$

    Thus, the simplified form is $4+0i$.