Write the first few values of the imaginary term $\displaystyle i$.

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

$\displaystyle i^2=i\cdot i = -1$

$\displaystyle i^4=i^2\cdot i^2 = (-1)\cdot (-1) =1$

Rewrite the expression using the product of exponents.

$\displaystyle -2i^{5000} = -2(i^4)^{1250}= -2(1)^{1250}$

The value of one to this power will remain the same.

$\displaystyle -2(1) =-2$

The answer is:  $\displaystyle -2$

To be able to evaluate the expression, we will need to write out the value of the imaginary terms.

Recall that:  $\displaystyle i=\sqrt{-1}$

$\displaystyle i^2= -1$

$\displaystyle i^4 = i^2\cdot i^2 = (-1)\cdot (-1) =1$

$\displaystyle i^6 = i^4 \cdot i^2 = (1) \cdot (-1) = -1$

Replace the terms.

$\displaystyle 2i^2+3i^4-4i^6-5i^2= 2(-1)+3(1)-4(-1)-5(-1)$

$\displaystyle =-2+3+4+5 = 10$

The answer is:  $\displaystyle 10$

Rewrite the problem as separate groups of binomials.

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

Use the FOIL method to expand the first two terms.

$\displaystyle (3+2i)(3+2i) = (3)(3)+(3)(2i)+(2i)(3)+(2i)(2i)$

Simplify the right side.

$\displaystyle 9+12i+4i^2$

Recall that since $\displaystyle i=\sqrt{-1}$, the value of $\displaystyle i^2 = i\cdot i =-1$.

$\displaystyle 9+12i+4(-1) = 12i+5$

Multiply this value with the third binomial.

$\displaystyle (12i+5)(3+2i) = (12i)(3)+(12i)(2i)+(5)(3)+(5)(2i)$

Simplify the terms.

$\displaystyle 36i+24i^2+15+10i=36i+24(-1)+15+10i = 46i-9$

The answer is:  $\displaystyle 46i-9$

In order to solve this expression, we will need to evaluate each term.

Use the property of exponents to subtract the powers in the fraction.

$\displaystyle \frac{-2i^3}{i^{-7}} = -2i^{3-(-7)} = -2i^{10}$

This term can be written as a fraction.

$\displaystyle -2i^{-10} = -2\cdot \frac{1}{i^{10}}$

Recall that the imaginary term $\displaystyle i=\sqrt{-1}$.  This means that:

$\displaystyle i^2 = i\cdot i = -1$

$\displaystyle i^{10} = (i^2)^5 = (-1)^5 = -1$

Replace the terms.

$\displaystyle -2\cdot \frac{1}{i^{10}} = -2\cdot \frac{1}{-1} = 2$

Solve the expression by replacing the values in the original expression.

$\displaystyle \frac{-2i^3}{i^{-7}}-i^2 = 2-(-1) = 2+1 =3$

The answer is:   $\displaystyle 3$

Write the first few powers of the imaginary term.

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

$\displaystyle i^2=i\cdot i = -1$

$\displaystyle i^3=i^2\cdot i = -i$

$\displaystyle i^4=i^2 \cdot i^2 = (-1)\cdot (-1) = 1$

We can then rewrite the higher powered imaginary terms by the product of exponents.

$\displaystyle -1000i^{52}+ 500i^{30} = -1000(i^4)^{13}+500(i^2)^{15}$

Simplify the terms.

$\displaystyle -1000(1)^{13}+500(-1)^{15} = -1000-500 = -1500$

The answer is:  $\displaystyle -1500$

Identify the first two powers of the imaginary term.

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

$\displaystyle i^2=i\cdot i =-1$

Rewrite the expression as a product of exponents.

$\displaystyle -5i^{1234}=-5(i^{2})^{617} = -5(-1)^{617}$

Negative one to an odd power will be negative one.

$\displaystyle -5(-1) =5$

The answer is:  $\displaystyle 5$

Write the first few powers of the imaginary term.

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

$\displaystyle i^2 = i\cdot i= -1$

Change the higher ordered power by using the power rule of exponents.

$\displaystyle 2i^{878} = 2(i^2)^{439}=2(-1)^{439}=2(-1)$

A negative one to an odd power will be negative one.

The answer is:  $\displaystyle -2$

The complex conjugate of a complex number $\displaystyle a+bi$ is $\displaystyle a-bi$. 

Therefore, the complex conjugate of $\displaystyle 8 +2 i \sqrt{5}$ is $\displaystyle 8 -2 i \sqrt{5}$. Add the two:

$\displaystyle \left (8 +2 i \sqrt{5} \right )+\left ( 8 -2 i \sqrt{5} \right )$

Collect real parts and imaginary parts:

$\displaystyle =8+ 8 +2 i \sqrt{5} -2 i \sqrt{5}$

The imaginary parts cancel out:

$\displaystyle =8+ 8$

$\displaystyle =16$

The complex conjugate of a complex number $\displaystyle a-bi$ is $\displaystyle a+bi$, so the complex conjugate of $\displaystyle 9 - i \sqrt{5}$ is $\displaystyle 9 + i \sqrt{5}$.

To raise $\displaystyle i$ to the power of a negative integer, first raise $\displaystyle i$ to the absolute value of that integer. Therefore, to find $\displaystyle i ^{-34}$, we need to look at $\displaystyle i^{34}$. 

To raise $\displaystyle i$ to the power of any positive integer, divide the integer by 4 and note the remainder. The correct power is given according to the table below.

$\displaystyle 34 \div 2 = 16 \textrm{ R }2$, so $\displaystyle i^{34}$ can be determined by selecting the power of $\displaystyle i$ corresponding to remainder 2. This is $\displaystyle -1$. 

Since $\displaystyle i^{34} = -1$, and by definition, $\displaystyle a^{-N} = \frac{1}{a^{N}}$, it follows that 

$\displaystyle i^{-34} = \frac{1}{i^{34} }= \frac{1}{-1} = -1$.