COMPLEX NUMBERS · HL

Complex Numbers 2 — Multiply and Conjugate

Multiplying complex numbers and the complex conjugate.

Section 1 of 3

Multiply

Method

1.Split the brackets.

Simplify

$(x + 5)(x + 7)$

$x(x + 7) + 5(x + 7)$

$x^2 + 7x + 5x + 35$

$x^2 + 12x + 35$

Same — split the brackets, but

1.$i^2 = -1$

(i) $(3 + i)(4 + 5i)$

$3(4 + 5i) + i(4 + 5i)$

$12 + 15i + 4i + 5i^2$

$12 + 19i + 5(-1)$

$12 + 19i - 5$

$7 + 19i$

(ii) $(3 + 5i)(6 + 4i)$

$= 3(6 + 4i) + 5i(6 + 4i)$

$= 18 + 12i + 30i + 20i^2$

$= 18 + 42i + 20(-1)$

$= 18 + 42i - 20$

$= -2 + 42i$

(iii) $(5 - 2i)(3 + 8i)$

$5(3 + 8i) - 2i(3 + 8i)$

$15 + 40i - 6i - 16i^2$

$15 + 34i - 16(-1)$

$15 + 34i + 16$

$31 + 34i$

(iv) $(2 - 5i)(3 - 7i)$

$2(3 - 7i) - 5i(3 - 7i)$

$6 - 14i - 15i + 35i^2$

$6 - 29i - 35$

$-29 - 29i$

(v) $(4 - 3i)(7 - 2i)$

$4(7 - 2i) - 3i(7 - 2i)$

$28 - 8i - 21i + 6i^2$

$28 - 29i - 6$

$22 - 29i$

Section 2 of 3

Simplify

Remember

1.$i^2 = -1$

(i) $(3 + 5i)(2 + 6i)$

$3(2 + 6i) + 5i(2 + 6i)$

$6 + 18i + 10i + 30i^2$

$6 + 28i - 30$

$-24 + 28i$

(ii) $(5 - 3i)(6 - 2i)$

$5(6 - 2i) - 3i(6 - 2i)$

$30 - 10i - 18i + 6i^2$

$30 - 28i - 6$

$24 - 28i$

(iii) $(3 - 5i)(7 - 3i)$

$3(7 - 3i) - 5i(7 - 3i)$

$21 - 9i - 35i + 15i^2$

$21 - 44i - 15$

$6 - 44i$

(iv) $(3 + 5i)^2$

Square means by itself.

$(3 + 5i)(3 + 5i)$

$3(3 + 5i) + 5i(3 + 5i)$

$9 + 15i + 15i + 25i^2$

$9 + 30i - 25 = -16 + 30i$

(x) $(5 + 7i)(5 - 7i)$

$5(5 - 7i) + 7i(5 - 7i)$

$25 - 35i + 35i - 49i^2$

$25 + 49 = 74$

Section 3 of 3

Conjugate

1.Every complex number $z = x + yi$ has a complex conjugate written $\bar{z} = x - yi$.

2.Change sign of the imaginary part.

$z = 5 + 3i$. Find:

$z = 5 + 3i, \qquad \bar{z} = 5 - 3i$

(i) $z + \bar{z}$

$5 + 3i + 5 - 3i$

$= 10$

(ii) $z \cdot \bar{z}$

$(5 + 3i)(5 - 3i)$

$5(5 - 3i) + 3i(5 - 3i)$

$25 - 15i + 15i - 9i^2$

$25 + 9 = 34$

SUM

The lot in one box

Multiply and Conjugate toolkit

1.Split the brackets, then use $i^2 = -1$.

2.Square means by itself: $(a + bi)^2 = (a + bi)(a + bi)$.

3.Conjugate: $\bar{z} = x - yi$ — change sign of the imaginary part.

End of lesson

Complex Numbers 2 — Multiply and Conjugate · HL · Mathslive.ie