Multiplication of Imaginaries

Multiplication of imaginaries

By the definition of square root, the square of \sqrt{-n} is -n. Therefore


\left(\sqrt{-1}\right)^2 = -1,	i^2 = -1
\left(\sqrt{-1}\right)^3 = \left(\sqrt{-1}\right)^2\sqrt{-1} = -1\sqrt{-1},	i^3 = -i
\left(\sqrt{-1}\right)^4 = \left(\sqrt{-1}\right)^2\left(\sqrt{-1}\right)^2 = (-1)(-1) = 1,	i^4 = 1

To multiply \sqrt{-2} by \sqrt{-3} we write \sqrt{-2} as \sqrt2 \cdot \sqrt{-1}, and \sqrt{-3} as \sqrt3 \cdot \sqrt{-1}. Then

\sqrt{-2} \cdot \sqrt{-3} = \left(\sqrt2 \cdot \sqrt{-1}\right)\left(\sqrt3 \cdot \sqrt{-1}\right) = \sqrt6 \cdot \sqrt{-1} \cdot \sqrt{-1} = -\sqrt6.

Similarly 2\sqrt{-5}\left(-3\sqrt{-2}\right) = 2\sqrt5 \cdot \sqrt{-1}\left(-3\sqrt2 \cdot \sqrt{-1}\right) = 6\sqrt{10}.

In general, if \sqrt{-a} and \sqrt{-b} are two imaginaries whose product (or quotient) is desired, they should first be written in the form \sqrt a \cdot \sqrt{-1} and \sqrt b \cdot \sqrt{-1} and the multiplication (or division) should then be performed. This method will prevent many errors.

In this connection it must be clearly understood that one rule followed in multiplication of radicals does not apply to imaginary numbers.

Thus \sqrt2 \cdot \sqrt3 = \sqrt{2 \cdot 3} = \sqrt6. But \sqrt{-2} \cdot \sqrt{-3} does not equal \sqrt{(-2)(-3)}, which equals \sqrt6.

In multiplying two complex numbers, first write each expression in the form a + bi, and then proceed as in the following

Example. Multiply 2 + \sqrt{-3} by 3 - \sqrt{-7}.


	2 + \sqrt{-3} = 2 + \sqrt3\sqrt{-1}
	3 - \sqrt{-7} = 3 - \sqrt7\sqrt{-1}
Multiplying,	6 + 3\sqrt3\sqrt{-1} - 2\sqrt7\sqrt{-1} + \sqrt{21}
Rewriting,	6 + \sqrt{21} + \left(3\sqrt3 - 2\sqrt7\right)i

(The last term is -\sqrt{21}\left(\sqrt{-1}\right)^2 = +\sqrt{21}.)

How to type imaginaries

Type the imaginary unit as i: 3i, 2 - 5i, sqrt(2) i, 2x i. You may also type it as sqrt(-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) becomes 3i, sqrt(-2) becomes sqrt(2) i.
A fraction may stand over its common denominator: (1 + 3i)/5 and 1/5 + 3/5 i are both fine.
Letters stand for positive numbers, so \sqrt{-a^2} = a\sqrt{-1} = ai.

From the history of algebra

Cardan and the first complex products. Long before the time of Gauss mathematicians had performed the operations of multiplication and division on complex numbers by the same rules that they used for real numbers. As early as 1545 Cardan stated that the product of 5 + \sqrt{-15} and 5 - \sqrt{-15} was 40. He was not always equally fortunate in obtaining correct results, for in another place he sets \frac14\left(-\sqrt{-\frac14}\right) = \frac{1}{\sqrt{64}} = \frac18.

Even the rather complicated formula for extracting any root of a complex number was discovered in the early part of the eighteenth century. But all of these operations were purely formal, and seemed to most mathematicians a mere juggling with symbols until Gauss showed clearly the place and usefulness of such numbers.

Exercises

0 of 25 solved

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 multiplications and simplify results:

\left(\sqrt{-1}\right)^5

\left(\sqrt{-1}\right)^6

\left(\sqrt{-1}\right)^7

\left(\sqrt{-1}\right)^8

2\sqrt{-1} \cdot 3\sqrt{-1}

\sqrt{-9} \cdot \sqrt{-16}

\sqrt{-5}\left(-\sqrt{-6}\right)

\sqrt{-25} \cdot \sqrt3

2\sqrt{-3} \cdot 3\sqrt{-2}

\sqrt{-m} \cdot \sqrt{-n}

4\sqrt{-5}\left(-3\sqrt{-6}\right)

\sqrt{a + b} \cdot \sqrt{-a - b}

\left(2 + \sqrt{-1}\right)\left(2 - \sqrt{-1}\right)

\left(3 + \sqrt{-2}\right)\left(3 - \sqrt{-2}\right)

\left(4 - 2\sqrt3\,i\right)\left(4 + 2\sqrt3\,i\right)

\left(3 + \sqrt{-1}\right)\left(6 - \sqrt{-2}\right)

(4 - 2i)\left(3 - 2\sqrt3\,i\right)

(a + ib)(c + id)

(a + ib)(a + ib)

(a + bi)(a - bi)

\left(-\tfrac12 + \tfrac12\sqrt{-3}\right)^2

\left(-\tfrac12 - \tfrac12\sqrt{-3}\right)^2

(x - iy)^2 - (x + iy)^2

\left(-\tfrac12 + \tfrac12\sqrt{-3}\right)^3 - \left(-\tfrac12 - \tfrac12\sqrt{-3}\right)^3

\left(a + i\sqrt{1 - x^2}\right)\left(a - i\sqrt{1 - x^2}\right)