By the power of a product property, 

$\displaystyle \left (3i \right )^{5} = 3^{5} \cdot i^{5} = 3^{5} \cdot i^{4} \cdot i = 243 \cdot 1 \cdot i = 243 i$

This is the product of a complex number and its complex conjugate. They can be multiplied using the pattern

$\displaystyle \left (A + Bi \right )\left (A - Bi \right ) = A^{2} + B^{2}$

with $\displaystyle A = 10, B = 3$

$\displaystyle (10 + 3i)(10-3i) = 10^{2} +3^{2} = 100 + 9 = 109$

$\displaystyle i ^{-56} = \frac{1}{i^{56}}$

$\displaystyle 56 \div 4 = 14$, so 56 is a multiple of 4. $\displaystyle i$ raised to the power of any multiple of 4 is equal to 1, so

$\displaystyle i ^{-56} = \frac{1}{i^{56}} = \frac{1}{1} = 1$.

The first step to solving this problem is distributing the exponent:

$\displaystyle 2^7i^7 = 128i^7$

Next, we need simplify the complex portion.

$\displaystyle i^7=i^2\times i^2\times i^2\times i=-1\times -1\times -1\times i=-i$

Thus, our final answer is $\displaystyle 128\times (-i)=-128i$.

$\displaystyle \left (3i \right )^{8}$

$\displaystyle = 3^{8} i^{8}$

$\displaystyle = 6,561 i^{8}$

$\displaystyle i$ raised to the power of any multiple of 4 is equal to 1, so the above expresion is equal to 

$\displaystyle = 6,561 \cdot 1 = 6,561$

This is not among the given choices.

You are being asked to evaluate

$\displaystyle \left (2 + i\sqrt{5} \right )^{3}$

You can use the cube of a binomial pattern with $\displaystyle A = 2, B = i\sqrt{5}$:

$\displaystyle \left ( A + B \right )^{3} = A^{3}+ 3A^{2}B + 3AB^{2}+ B^{3}$

$\displaystyle \left ( 2 + i\sqrt{5} \right )^{3} =2^{3}+ 3 \cdot 2^{2} \cdot i\sqrt{5} + 3 \cdot 2 (i\sqrt{5}) ^{2}+ (i\sqrt{5}) ^{3}$

$\displaystyle =8+ 3 \cdot 4 \cdot i\sqrt{5} + 6i ^{2} (\sqrt{5}) ^{2}+ i^{3}(\sqrt{5}) ^{3}$

$\displaystyle =8+12i\sqrt{5} + 6 (-1) 5+(-i)(5\sqrt{5})$

$\displaystyle =8-30 +12i\sqrt{5} -5i\sqrt{5}$

$\displaystyle =-22 +7i\sqrt{5}$

$\displaystyle \left (3 - i\sqrt{2} \right )^{4}$ can be calculated by squaring $\displaystyle 3 - i\sqrt{2}$, then squaring the result, using the square of a binomial pattern as follows:

$\displaystyle \left (3 - i\sqrt{2} \right )^{2}$

$\displaystyle =3^{2}- 2 \cdot 3 \cdot i\sqrt{2} +\left ( i\sqrt{2} \right )^{2}$

$\displaystyle =9- 6 i\sqrt{2} +2i^{2}$

$\displaystyle =9- 6 i\sqrt{2} +2(-1)$

$\displaystyle =7- 6 i\sqrt{2}$

$\displaystyle \left (3 - i\sqrt{2} \right )^{4}$

$\displaystyle =\left [ \left (3 - i\sqrt{2} \right )^{2} \right ]^{2}$

$\displaystyle =\left (7- 6 i\sqrt{2}} \right )^{2}$

$\displaystyle =7^{2} - 2 \cdot 7 \cdot 6 i\sqrt{2}} + \left ( 6 i\sqrt{2}} \right )^{2}$

$\displaystyle =49-84 i\sqrt{2}} + 6^{2} i^{2} \left ( \sqrt{2}} \right )^{2}$

$\displaystyle =49-84 i\sqrt{2}} +36 (-1) (2)$

$\displaystyle =49-84 i\sqrt{2}} -72$

$\displaystyle =-23-84 i\sqrt{2}}$

The product of any complex number $\displaystyle a + bi$ and its complex conjugate $\displaystyle a - bi$ is the real number $\displaystyle a^{2}+b^{2}$, so all that is needed here is to evaluate the expression:

$\displaystyle 32^{2}+ 18 ^{2}= 1,024+ 324 = 1,348$

First, square $\displaystyle 1+ i$ using the square of a binomial pattern as follows:

$\displaystyle \left (1+ i \right )^{2}$

$\displaystyle = 1^{2}+ 2 \cdot 1 \cdot i + i^{2}$

$\displaystyle = 1 +2i -1$

$\displaystyle = 2i$

Raising this number to the fourth power yields the correct response:

$\displaystyle \left (1+ i \right )^{8}$

$\displaystyle = \left [\left (1+ i \right )^{2} \right ]^{4}$

$\displaystyle = (2i)^{4}$

$\displaystyle = (2 )^{4} \cdot i^{4}$

$\displaystyle =16 \cdot 1 = 16$

$\displaystyle \left ( -2i\right )^{9}$

$\displaystyle = \left ( -2 \right )^{9} \cdot i^{9}$

To raise a negative number to an odd power, take the absolute value of the base to that power and give its opposite:

$\displaystyle \left ( -2\right )^{9} = - \left ( 2^{9}\right ) = -512$

To raise $\displaystyle i$ to a power, divide the power by 4 and raise $\displaystyle i$ to the remainder. Since 

$\displaystyle 9 \div 4 = 2 \textup{ R }1$,

$\displaystyle i^{9} = i^{1}= i$

Therefore, 

$\displaystyle \left ( -2i\right )^{9}= \left ( -2 \right )^{9} \cdot i^{9} = -512 i$