# Throughout the semester- we were not allowed to take square roots of n.docx

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Throughout the semester, we were not allowed to take square roots of negative numbers. The reason why this was so is because we never introduced the set of complex numbers, usually denoted by C. Numbers in this set are of the form a + ib, where a and b are real numbers. For now, you can think of i as a symbol. i has the property that i2 = ??1, which is equivalent to i = p ??1. For example, one can show that (a + ib)2 = a2 ?? b2 + 2iab and (a + ib)(a ?? ib) = a2 + b2 (work it out). Let a + ib be a complex number. The complex conjugate is the number a ?? ib. Division by complex numbers doesn\'t behave easy. The number a + ib c + id is NOT in the form A + iB because there is an i in the denominator. However, we can always turn a number of this form into a complex number of the form A + iB. Show how to do this. Here\'s a hint: multiply both the numerator and denominator of a + ib c + id by the complex conjugate of the denominator. This amounts to multiplying the entire fraction by 1 because you multiplied the top and bottom by the same thing. After some manipulation, you will get a number of the form A + iB. Solution (a+ib)x(c-id)/(c+id)x(c-id) =(ac+bd + i(bc-ad))/(c^2 +d^2)= A+iB (ac+bd)/(c^2 + d^2) =A i(bc-ad)/(c^2 + d^2) = iB .

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### Throughout the semester- we were not allowed to take square roots of n.docx

• 1. Throughout the semester, we were not allowed to take square roots of negative numbers. The reason why this was so is because we never introduced the set of complex numbers, usually denoted by C. Numbers in this set are of the form a + ib, where a and b are real numbers. For now, you can think of i as a symbol. i has the property that i2 = ??1, which is equivalent to i = p ??1. For example, one can show that (a + ib)2 = a2 ?? b2 + 2iab and (a + ib)(a ?? ib) = a2 + b2 (work it out). Let a + ib be a complex number. The complex conjugate is the number a ?? ib. Division by complex numbers doesn't behave easy. The number a + ib c + id is NOT in the form A + iB because there is an i in the denominator. However, we can always turn a number of this form into a complex number of the form A + iB. Show how to do this. Here's a hint: multiply both the numerator and denominator of a + ib c + id by the complex conjugate of the denominator. This amounts to multiplying the entire fraction by 1 because you multiplied the top and bottom by the same thing. After some manipulation, you will get a number of the form A + iB. Solution (a+ib)x(c-id)/(c+id)x(c-id) =(ac+bd + i(bc-ad))/(c^2 +d^2)= A+iB (ac+bd)/(c^2 + d^2) =A i(bc- ad)/(c^2 + d^2) = iB
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