Nothing can be simplified in either parentheses, so the first step is to distribute the negative sign to the second parentheses

$\displaystyle 3i+2-6i-5$

Then, you combine similar terms remembering that terms with i cannot combine with those with no i

$\displaystyle 9i-3$

In order to simplify this, we will need to identify the solution to each term with an imaginary power.

Recall that $\displaystyle i=\sqrt{-1}$.  Write the solutions for the powers given.

$\displaystyle i^2= -1$

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

$\displaystyle i^5 = i^4\cdot i = i$

Rewrite and replace each term with the simplified forms.

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

Combine like-terms.

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

The answer is:  $\displaystyle -2i-4$

This problem should be foiled. Multiply the first two numbers of each parentheses, the outer numbers of the parentheses, the inner numbers of the two parentheses, and the last numbers of the parentheses.

$\displaystyle 10i^{2}-4i+15i-6$

Combine like terms

$\displaystyle 10i^{2}+11i-6$

Simplify the imaginary numbers. I squared equals -1

$\displaystyle 10(-1)+11i-6$

$\displaystyle -10+11i-6$

$\displaystyle -16+11i$

First perform the multiplication on the coefficients and the i's

$\displaystyle 3i\cdot -2i\cdot 5i$

$\displaystyle -30i^{3}$

i cubed is equal to negative i

$\displaystyle -30(-i)$

$\displaystyle 30i$

Multiply the coefficients and the i's together

$\displaystyle 5i\cdot i\cdot 4i\cdot -3i$

$\displaystyle -60i^{4}$

i to the fourth power is the same as the number one.

$\displaystyle -60(1)$

$\displaystyle -60$

This problem should be foiled. Multiply the first numbers of the parentheses, the outer numbers of the parentheses, the inner numbers of the parentheses, and the last numbers of the parentheses

$\displaystyle (11i-2)(10i-5)$

$\displaystyle 110i^{2}-55i-20i+10$

Combine like terms

$\displaystyle 110i^{2}-75i+10$

Simplify i squared to negative one

$\displaystyle 110(-1)-75i+10$

$\displaystyle -110-75i+10$

$\displaystyle -100-75i$

This problem can either be solved by simplifying the numbers in the parentheses first or by distributing the 8i to all terms in the parentheses. I will show simplifying the parentheses first.

Remember that i squared can be simplified to negative one and like terms can be combined.

$\displaystyle 8i(10i+5-4i+2i^{2})$

$\displaystyle 8i(6i+5+2(-1))$

$\displaystyle 8i(6i+5-2)$

$\displaystyle 8i(6i+3)$

Now distribute the 8i to both terms of the parentheses.

$\displaystyle 48i^{2}+24i$

Again, i squared can be changed to negative one

$\displaystyle 48(-1)+24i$

$\displaystyle -48+24i$

Nothing can be simplified in the parentheses but the sign outside of each parentheses can be distributed to each number inside.

$\displaystyle (6i-7)-(15i+2)+(10i-4)$

$\displaystyle 6i-7-15i-2+10i-4$

Combine like terms to get the answer

$\displaystyle i-13$

This problem should be foiled. Multiply the first numbers, the outer numbers, the inner numbers, and the last numbers of the parentheses.

$\displaystyle (-7i-15)(2+i)$

$\displaystyle -14i-7i^{2}-30-15i$

Combine like terms

$\displaystyle -29i-7i^{2}-30$

Simplifying i squared to negative one

$\displaystyle -29i-7(-1)-30$

$\displaystyle -29i+7-30$

$\displaystyle -29i-23$

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

Write the first few terms of the powers of the imaginary number.

$\displaystyle i^2 = -1$

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

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

$\displaystyle i^5 = i^4\cdot i = i$

Notice that the pattern will repeat itself for higher powers.

Rewrite the expression given as a product of exponents.

$\displaystyle i^{2000}-i^{1998} =(i^4)^{500}- (i^2)^{999}$

Simplify using the known values of $\displaystyle i^4$ and $\displaystyle i^2$.

$\displaystyle (i^4)^{500}- (i^2)^{999} = 1^{500}-(-1)^{999}$

The quantity of a negative number to an odd power will result in a negative number.  

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

The answer is:  $\displaystyle 2$