Simplifying i's
i3
What is -i
√(-81)
9i
(2 + 3i) + (4 + 7i)
6+10i
3i(2i2 - 5i)
15 - 6i
the square root of -1
i
i4
What is 1
√(-4) + √(-9)
5i
(5 + 6i) + (4 - 2i)
What is 9 + 4i
(6 + 2i)(6 - 2i)
40
Why do we need imaginary numbers?
To find the square root of negative numbers
i2
-1
√(-18)
3i√2
(6 + 5i) - (3 + 2i)
What is 3+3i
(2 + 3i)(4 + 7i)
-13+26i
Simplify the following: √(-36)
6i
i8
1
√(-32)
4i√2
(2 - 7i) - (1 - 4i)
What is 1-3i
(7 + i)(7 - i)
50
√(36) + √(-36)
6 + 6i
i19
-i
2√(-50)
10i√2
9 + 2i - (6 + 4i)
3 - 2i
3i(-8i + 5) + 2(5 - i)
34 + 13i
Why do we divide by 4 when simplifying powers of i?
Because the pattern repeats every 4.