The complex conjugate of a complex number $\displaystyle a+bi$ is $\displaystyle a - bi$, so $\displaystyle \frac{3}{2} + \frac{1}{4} i$ has $\displaystyle \frac{3}{2} - \frac{1}{4} i$ as its complex conjugate. Subtract the latter from the former:

$\displaystyle \left (\frac{3}{2} + \frac{1}{4} i \right ) - \left ( \frac{3}{2} - \frac{1}{4} i \right )$

$\displaystyle = \frac{3}{2} + \frac{1}{4} i - \frac{3}{2}+ \frac{1}{4} i$

$\displaystyle = \frac{3}{2}- \frac{3}{2} + \frac{1}{4} i + \frac{1}{4} i$

$\displaystyle = \frac{1}{2} i$

The complex conjugate of a complex number $\displaystyle a - bi$ is $\displaystyle a+ bi$. Therefore, the complex conjugate of $\displaystyle 17 - i \sqrt{17}$ is $\displaystyle 17 + i \sqrt{17}$; subtract the latter from the former by subtracting real parts and subtracting imaginary parts, as follows:

$\displaystyle (17 - i \sqrt{17})- (17 + i \sqrt{17})$

$\displaystyle = 17 - i \sqrt{17} - 17 - i \sqrt{17}$

$\displaystyle = 17 - 17 - i \sqrt{17} - i \sqrt{17}$

$\displaystyle = - 2 i \sqrt{17}$

The complex conjugate of a complex number $\displaystyle a+ bi$ is $\displaystyle a - bi$. Therefore, the complex conjugate of $\displaystyle 7 + i \sqrt {7 }$ is $\displaystyle 7 - i \sqrt {7 }$; subtract the latter from the former by subtracting real parts and subtracting imaginary parts, as follows:

$\displaystyle (7 + i \sqrt {7 })- (7 - i \sqrt {7 })$

$\displaystyle = 7- 7 + i \sqrt {7 }- (- i \sqrt {7 })$

$\displaystyle = i \sqrt {7 }+ i \sqrt {7 }$

$\displaystyle =2i \sqrt {7 }$

Rewrite $\displaystyle \sqrt{-3}+\sqrt{-9}+\sqrt{-16}$ in their imaginary terms.

$\displaystyle i=\sqrt{-1}$

$\displaystyle \sqrt{-3}+\sqrt{-9}+\sqrt{-16}=i\sqrt3+3i+4i = i\sqrt3 +7i$

The complex conjugate of a complex number $\displaystyle a+ bi$ is $\displaystyle a - bi$. Therefore, the complex conjugate of $\displaystyle 5 + i \sqrt {5 }$ is $\displaystyle 5 - i \sqrt {5 }$; add them by adding real parts and adding imaginary parts, as follows:

$\displaystyle (5 + i \sqrt {5 } )+( 5 - i \sqrt {5 })$

$\displaystyle = 5 + 5 + i \sqrt {5 } - i \sqrt {5 }$

$\displaystyle = 10$,

the correct response.

The complex conjugate of a complex number $\displaystyle a - bi$ is $\displaystyle a+ bi$. Therefore, the complex conjugate of $\displaystyle 13 - i \sqrt{13}$ is $\displaystyle 13 + i \sqrt{13}$; add them by adding real parts and adding imaginary parts, as follows:

$\displaystyle (13 - i \sqrt{13} )+( 13 + i \sqrt{13})$

$\displaystyle (13 - i \sqrt{13} )+( 13 + i \sqrt{13})$

$\displaystyle = 13+13 - i \sqrt {13 }+ i \sqrt {13}$

$\displaystyle = 26$

The common difference $\displaystyle d$ of an arithmetic sequence can be found by subtracting the first term from the second:

$\displaystyle d = a_{2} - a_{1}$

$\displaystyle d = 3i - 3 = -3 + 3i$

Add this to the second term to obtain the desired third term:

$\displaystyle a_{3} = a_{2} +d = 3i + ( -3 + 3i) = -3 + 3i + 3i = -3 + 6i$.

It can be easier to line real and imaginary parts vertically to keep things organized, but in essence, combine like terms (where 'like' here means real or imaginary):

$\displaystyle (3+5i)+(4-2i)+(-2+i)$

$\displaystyle =3+4+(-2)+5i+(-2i)+i$

$\displaystyle =5+4i$

The complex conjugate of a complex number $\displaystyle a+bi$ is $\displaystyle a - bi$, so $\displaystyle \frac{3}{2} + \frac{1}{4} i$ has $\displaystyle \frac{3}{2} - \frac{1}{4} i$ as its complex conjugate. The sum of the two numbers is

$\displaystyle \left (\frac{3}{2} + \frac{1}{4} i \right ) + \left ( \frac{3}{2} - \frac{1}{4} i \right )$

$\displaystyle = \frac{3}{2} + \frac{1}{4} i + \frac{3}{2} - \frac{1}{4} i$

$\displaystyle = \frac{3}{2} + \frac{3}{2} + \frac{1}{4} i - \frac{1}{4} i$

$\displaystyle =3$

A power of $\displaystyle i$ can be evaluated by dividing the exponent by 4 and noting the remainder. The power is determined according to the following table:

$\displaystyle \begin{matrix} \textrm{\underline{Rem}}& \textrm{\underline{Power}} \\ 0& 1\\ 1& i\\ 2& -1\\ 3& -i \end{matrix}$

$\displaystyle 9 \div 4 = 2 \textrm{ R }1$, so $\displaystyle i^{9} = i$

$\displaystyle 10 \div 4 = 2 \textrm{ R }2$, so $\displaystyle i^{10} = -1$

$\displaystyle 11 \div 4 = 2 \textrm{ R }3$, so $\displaystyle i^{11} = - i$

$\displaystyle 12 \div 4 = 3 \textrm{ R }0$, so $\displaystyle i^{12} = 1$

Substituting:

$\displaystyle 9i^{9}+ 10 i^{10}+ 11 i^{11}+ 12 i^{12}$

$\displaystyle = 9(i) + 10 (-1)+ 11 (-i)+ 12 (1)$

$\displaystyle = 9i - 10 - 11i+ 12$

Collect real and imaginary terms:

$\displaystyle - 10 + 12 + 9i - 11i$

$\displaystyle = 2 - 2i$