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# 15_Complex Numbers

### Terms and Practice of Complex Numbers

Question | Answer |
---|---|

Write √(-16) using i notation. | √(-16) = √(-1 x 16) = √(-1) x √(16)= ix4, or 4i |

Multiply √(-17) x √(-14). | √(-17) x √(-14) = i√(17)x i√(14) = i^2√(238) = -1√(238) = -√(238) |

Divide √(-27)/√(3). | √(-27)/√(3) = i√(27)/√(3) = i√(9) = 3i |

Add (2+3i)+(-3+2i). Write in a+bi form. | (2+3i)+(-3+2i)= (2-3)+(3+2)i = -1+5i |

Subtract (-3-7i)-(-6). Write in a+bi form. | (-3-7i)-(-6)= -3-7i+6 = (-3+6)-7i = 3-7i |

Multiply 3i(2-i). Write in a+bi form. | 3i(2-i)= 3i(2)-3i(i) = 6i-3i^2 = 6i-3(-1) = 6i+3 = 3+6i |

Multiply (7+3i)(7-3i). Write in a+bi form. | (7+3i)(7-3i) = 7(7)-7(3i)+3i(7)-3i(3i) = 49-21i +21i -9i^2 = 49-9(-1) = 49+9 = 58 |

Divide (2+i)/(1-i). Write in a+bi form. | (2+i)/(1-i)= (2+i)(1+i)/(1-i)(1+i) = 2(1)+2(i)+1(i)+i^2/(1^2-i^2) = (2+3i-1)/(1+1) = (1+3i)/2, or (1/2)+(3/2)i |

Find the power of i^6. | i^6 = (i^4)(i^2) = 1(-1) = -1 |

Find the power of i^22 | i^22 = (i^20)(i^2) = (i^4)^5(i^2) = 1^5(-1) = -1 |

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