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:

\(\displaystyle - 10 + 12 + 9i - 11i\)

\(\displaystyle = 2 - 2i\)

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:

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

\(\displaystyle 6 \div 4 = 1 \textrm{ R }2\), so \(\displaystyle i^{6} = -1\)

\(\displaystyle 7 \div 4 = 1 \textrm{ R }3\), so \(\displaystyle i^{7} = -i\)

\(\displaystyle 8 \div 4 = 2 \textrm{ R }0\), so \(\displaystyle i^{8} = 1\)

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

Substituting: 

\(\displaystyle i^{6} + i^{7}+ i^{8}+ i^{9}\)

\(\displaystyle = -1+ ( - i )+ 1 +i\)

\(\displaystyle = -1 + 1 - i + i\)

\(\displaystyle = 0\)

 \(\displaystyle {x^{-\frac{1}{2}}}\)  is the same as \(\displaystyle \frac{1}{\sqrt{x}}\) , which means that the bigger the answer to \(\displaystyle \sqrt{x}\) is, the smaller the fraction will be.

Therefore, \(\displaystyle 16\) is the correct answer because 

\(\displaystyle \frac{1}{\sqrt{16}} = \frac{1}{4}\).