Simplify.

1 + i² + 3i⁵ + 8i¹³ + 21i³⁴

Hmm, this sequence of numbers looks familiar. Fibonacci, is that you?

When you have an exponent on an i, check to see if it's 4 or less. If so, simplify. If not, find as many powers of 4 as you can, because each i⁴ equals 1.

1 + (-1) + 3i(i⁴) + 8i(i⁴)³ + 21i²(i⁴)⁸

Nice. Now replace every stinkin' i⁴ with a nice, easy 1 and simplify.

0 + 3i(1) + 8i(1) + 21(-1)(1) =
3i + 8i – 21 =
-21 + 11i

Remember to put the real number portion first and the imaginary part second. Them's the rules.

Simplify.

(4i + 3)(-2i + 1)(1 – i)²

First we bust out the distributive property on the first two sets of parentheses, then do the same thing with the squared stuff.

(-8i² + 4i – 6i + 3)(1 – 2i + i²)

Now we combine like terms and replace any stray i² with a -1.

(-8(-1) – 2i + 3)(1 – 2i + (-1)) =

(11 – 2i)(-2i) =

-22i + 4i² =

-4 – 22i

Simplify.

5i⁶ + 3 + (2i² – 1)(3i – 7)

You can start off by simplifying some of the imaginary terms.

5i²(i⁴) + 3 + (2(-1) – 1)(3i – 7) =

5(-1)(1) + 3 + (-2 – 1)(3i – 7) =

-2 – 3(3i – 7)

Now distribute that -3 like a boss.

-2 – 9i + 21

19 – 9i

See? Equations with complex numbers usually aren't as complex as they seem.