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# Module 15

### Complex Numbers

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

Write the the square root of -9 in terms of i | To write the square root of a negative number in terms of i, use the property that if a is a positive number then √-a = √-1 * √a =i√a. so using i we can write √ -9 = √ -1*9 = √-1 * √9= i*3 or 3i |

Write with i notation √-4 | √-4= √-1*4= √-1*√4= 2i |

Write with i notation √-7 | √-1*7 = √-1*√7 =i√7 |

Write with i notation -√8 | -√8 = -√-1*8 = √-1* √4*2 = -i*2√2= -2i√2 |

Multiply √-9 and √-16 | We must first write each number in the form of bi. √-9√-16= 3i*4i = 12i^2= 12(-1) = -12 |

Multiply √-6*√-2 | √-6*√-2 = i√6(i√2)=i^2√12 = -1(2)√3= -2√3 |

Multiply √4*√-4 | √4*√-4 = 2(√-1*√4)= 2(2i)= 4i |

Divide √-32/√8 | √-32/√8 = i√32/√8= i√4= 2i |

Add the complex number (4+3i)+(-3+4i)write the sum in the form a + bi | (4+3i)+(-3+4i) = (4-3)+(4+3)i = 1+7i |

Subtract the complex number 7i-(1-i)write the difference in the form a+bi | 7i-(1-i)=7i-1+i= -1 +(7+1)i = -1+8i |

Multiply the complex number -6i* 2i write the product in the form a+bi | -6i*2i= -12i^2 = -12(-1)= 12+0i |

Multiply the complex number (3-i)^2 | (3-i)^2 = (3-i)(3-i)= 3(3)-3(i)-3(i)+i^2= 9-6i+(-1)= 9-6i |

Divide 5/2i | Multiply the numerator and denominator by the conjugate of 2i. 2i = 0+2i so its conjugate is 0-2i or -2i. 5/2i= 5(-2i)/2i(-2i)= -10i/-4i^2= -10i/-4(-1)= -10i/4= -5i/2 or 0-(5/2)i |

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