Lesson 3 · from §119 of the 1910 text
Division of Imaginaries
Division of imaginaries
One complex number is the conjugate of another if their product is real. Thus a + bi and a - bi are conjugates. Conjugate complex numbers are used in division of imaginary expressions as conjugate radicals are used in division of radicals.
Division of imaginaries is performed by writing the dividend over the divisor as a fraction and then multiplying both numerator and denominator by the simplest imaginary expression which will make the resulting denominator real and rational.
Example 1. \sqrt{-6} \div \sqrt{-2}.
\sqrt{-6} \div \sqrt{-2} = \frac{\sqrt{-6}}{\sqrt{-2}} = \frac{\sqrt{-6} \cdot \sqrt{-2}}{\sqrt{-2} \cdot \sqrt{-2}} = \frac{\sqrt6 \cdot \sqrt{-1} \cdot \sqrt2 \cdot \sqrt{-1}}{\sqrt2 \cdot \sqrt{-1} \cdot \sqrt2 \cdot \sqrt{-1}} = \frac{-\sqrt{12}}{-2} = \sqrt3.
Example 2. 3 \div \left(2 + \sqrt{-3}\right).
3 \div \left(2 + \sqrt{-3}\right) = \frac{3}{2 + \sqrt{-3}} = \frac{3\left(2 - \sqrt{-3}\right)}{\left(2 + \sqrt{-3}\right)\left(2 - \sqrt{-3}\right)} = \frac{6 - 3\sqrt{-3}}{4 + 3} = \frac{6 - 3\sqrt{-3}}{7}.
In the form a + bi this is \frac67 - \frac37\sqrt3\,i.
A real number under a fourth root of a negative number works the same way: \sqrt[4]{-16} = \sqrt[4]{16}\,\sqrt[4]{-1} = 2\sqrt[4]{-1}, so in a quotient of two such roots the factors \sqrt[4]{-1} cancel.
How to type imaginaries
- Type the imaginary unit as
i:3i,2 - 5i,sqrt(2) i,2x i. You may also type it assqrt(-1), the way the book writes it:5sqrt(-1). - Give each answer in the form a + bi: work out every power of i, collect the real terms and the imaginary terms, and leave no i in a denominator. Write the square root of a negative number as a multiple of i:
sqrt(-9)becomes3i,sqrt(-2)becomessqrt(2) i. - A fraction may stand over its common denominator:
(1 + 3i)/5and1/5 + 3/5 iare both fine. - Letters stand for positive numbers, so \sqrt{-a^2} = a\sqrt{-1} = ai.
Real answers are typed as usual, in simplest radical form with no radical in a denominator: 2sqrt(3)/3, (a - 1) root(a, 4).
Exercises
How to type math
Use ^ for powers (a^2), / for fractions ((a+b)/2), sqrt(x) for square roots, and cbrt(x) for cube roots. You can leave out the multiplication sign: 3ab means 3 \cdot a \cdot b. Mixed numbers like 4 1/2 work too. The preview under each box shows how your answer was read.
Perform the indicated division:
\sqrt{-8} \div \sqrt{-2}
\sqrt{-6} \div \sqrt{-2}
2\sqrt{-3} \div 3\sqrt{-1}
\sqrt[4]{-1} \div \sqrt[4]{-4}
\sqrt6 \div \sqrt{-2}
(-25)^{\frac14} \div (-81)^{\frac14}
\sqrt{ax} \div \sqrt{-x}
\sqrt{-a} \div \sqrt{-b}
(-5ax)^{\frac12} \div (-2x)^{\frac12}
\left[\left(-a^6\right)^{\frac14} - \left(-a^2\right)^{\frac14}\right] \div (-a)^{\frac14}
2 \div \left(1 - \sqrt{-1}\right)
3 \div \left(2 - \sqrt{-2}\right)
2\sqrt{-1} \div \left(\sqrt{-1} + 3\right)
3\sqrt{-2} \div \left(2\sqrt{-3} + 2\right)
\dfrac{-1 + \sqrt{-3}}{-1 - \sqrt{-3}}
\dfrac{1 + i}{1 - i}
\dfrac{a}{a + bi}
\dfrac{a + ib}{c + id}
Does x^2 - 4x + 7 = 0, if x = 2 \pm \sqrt{-3}?
Determine whether the sum and the product of 2 + 3\sqrt{-1} and 2 - 3\sqrt{-1} are real numbers: find each.
Show that the sum (and the product) of any two conjugate complex numbers is real: find the sum and the product of a + bi and a - bi.
Show that the quotient of two conjugate complex numbers is complex: find (a + bi) \div (a - bi) in the form \dfrac{p + qi}{r}.
Point out the error in the following:
| (1) Obviously | \sqrt{-1} = \sqrt{-1}. |
| (2) Then | \sqrt{\dfrac{-1}{1}} = \sqrt{\dfrac{1}{-1}}, |
| (3) or | \dfrac{\sqrt{-1}}{\sqrt1} = \dfrac{\sqrt1}{\sqrt{-1}}. |
| (4) Therefore | \left(\sqrt{-1}\right)^2 = \left(\sqrt1\right)^2. |
| (5) That is, | -1 = 1. |
Which line is the first wrong step?