Complex numbers Math 3 Honors

solve 4 81 16 144 −4 * −16 * ±2 ±9 ±4 ±12 ±2𝑖 ±4𝑖

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𝑖= −1 So, what else do you know? 𝑖 2 =−1 𝑖 3 =−𝑖 𝑖 4 =1

simplify (HINT: Divide by 4 and use remainder to simplify) 𝑖 23 𝑖 2006 𝑖 37 𝑖 828 −𝑖 −1 𝑖 1

Complex numbers have real and imaginary part 𝑎+𝑏𝑖

Add and subtract 7𝑖+9𝑖 −5+6𝑖 +(2−11𝑖) 2+3𝑖 − 4+2𝑖 −3+4𝑖 − 1+3𝑖 3−4𝑖 + 2−7𝑖 −(5−12𝑖) 16𝑖 −3−5𝑖 −2+𝑖 −4+𝑖 𝑖

multiply (2+5𝑖)(7+2𝑖) (7−4𝑖)(3+4𝑖) (2+3𝑖)(14+8𝑖) (5−4𝑖)(−11+15𝑖) 4+39𝑖 37+16𝑖 4+58𝑖 5+119𝑖

To divide: Multiply both parts by the conjugate of the bottom **Conjugate of (a+bi) = (a-bi)

Why must we do that? We can’t have 𝑖 in the denominator and the conjugate will cancel out the middle term for us!

Divide −3+𝑖 5−2𝑖 (60+90𝑖)÷(14+8𝑖) (−33−56𝑖) (5−12𝑖) (−63+23𝑖)÷(−11+15𝑖) −17−𝑖 29 6+3𝑖 3−4𝑖 3+2𝑖