Worked examples · Sheet 1

Operations and quotients in algebraic form

Theory used: Sum, product, reciprocal and quotient Conjugate and its properties Real part and imaginary part

Exercise 1

Write in algebraic form

z = (3 + 2 i) (1 - i) + (2 - i)^{2} .

(3 + 2 i) (1 - i) = 3 - 3 i + 2 i - 2 i^{2} = 5 - i

(2 - i)^{2} = 4 - 4 i + i^{2} = 3 - 4 i

z = (5 - i) + (3 - 4 i) = 8 - 5 i

The first product is expanded with the distributive property: the four terms $3$ , $- 3 i$ , $2 i$ , $- 2 i^{2}$ appear. Since $i^{2} = - 1$ , the term $- 2 i^{2}$ equals $2$ , so the real part is $3 + 2 = 5$ and the imaginary part is $- 3 + 2 = - 1$ .

The square is computed with the formula for the square of a binomial: $(2 - i)^{2} = 4 - 4 i + i^{2}$ . Since $i^{2} = - 1$ , the real part equals $4 - 1 = 3$ .

In the final sum the real parts are added together and the imaginary parts are added together, therefore $z = 8 - 5 i$ .

Exercise 2

Write in algebraic form

z = \frac{1 + 2 i}{3 - i} .

z = \frac{( 1 + 2 i ) ( 3 + i )}{( 3 - i ) ( 3 + i )} = \frac{3 + i + 6 i + 2 i ^{2}}{9 + 1} = \frac{1 + 7 i}{10} = \frac{1}{10} + \frac{7}{10} i

To eliminate the imaginary unit from the denominator, numerator and denominator are multiplied by $3 + i$ , the conjugate of the denominator. In the denominator the product $(3 - i) (3 + i) = 3^{2} + 1^{2} = 10$ appears, which is real because the product of a number and its conjugate is the square of the modulus.

In the numerator the product $(1 + 2 i) (3 + i)$ gives $3 + i + 6 i + 2 i^{2}$ ; since $2 i^{2} = - 2$ , the real part equals $3 - 2 = 1$ and the imaginary part equals $1 + 6 = 7$ . Dividing by $10$ separates the real part and the imaginary part of the result.

Exercise 3

Determine the real part and the imaginary part of

z = \frac{1 + i}{1 - i} + \frac{2}{i} .

\frac{1 + i}{1 - i} = \frac{( 1 + i ) ^{2}}{( 1 - i ) ( 1 + i )} = \frac{1 + 2 i + i ^{2}}{2} = \frac{2 i}{2} = i

\frac{2}{i} = \frac{2 \cdot ( - i )}{i \cdot ( - i )} = \frac{- 2 i}{1} = - 2 i

z = i - 2 i = - i, Re (z) = 0, Im (z) = - 1

In the first quotient numerator and denominator are multiplied by $1 + i$ , the conjugate of the denominator. In the numerator $(1 + i)^{2} = 1 + 2 i + i^{2}$ appears; since $i^{2} = - 1$ , the real part cancels and $2 i$ remains. In the denominator the product equals $1^{2} + 1^{2} = 2$ , so the quotient is $i$ .

In the second quotient the conjugate of $i$ is $- i$ , and the product $i \cdot (- i) = - i^{2} = 1$ makes the denominator real; the quotient equals $- 2 i$ .

The sum of the two results is $- i$ , a purely imaginary number, therefore the real part is $0$ and the imaginary part is $- 1$ .