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Number and Algebra - Complex numbers – basic operations (HL)

Grade 11IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A complex number is defined as z=a+biz = a + bi, where aa and bb are real numbers (a,b∈Ra, b \in \mathbb{R}) and ii is the imaginary unit.

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The imaginary unit ii is defined by the property i2=−1i^2 = -1. Consequently, −n=in\sqrt{-n} = i\sqrt{n} for n>0n > 0.

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For a complex number z=a+biz = a + bi, aa is the real part, denoted Re(z)\text{Re}(z), and bb is the imaginary part, denoted Im(z)\text{Im}(z).

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Two complex numbers are equal, a+bi=c+dia + bi = c + di, if and only if their real parts are equal (a=ca = c) and their imaginary parts are equal (b=db = d).

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The complex conjugate of z=a+biz = a + bi is zˉ=a−bi\bar{z} = a - bi. A key property is that the product zzˉz\bar{z} is always a non-negative real number: a2+b2a^2 + b^2.

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Addition and subtraction are performed by combining the real parts and imaginary parts separately: (a+bi)±(c+di)=(a±c)+(b±d)i(a + bi) \pm (c + di) = (a \pm c) + (b \pm d)i.

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Multiplication is performed using the distributive law (FOIL), replacing i2i^2 with −1-1.

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Division of complex numbers z1z2\frac{z_1}{z_2} is performed by multiplying both the numerator and the denominator by the conjugate of the denominator, zˉ2\bar{z}_2, to realize the denominator.

📐Formulae

i1=i,i2=−1,i3=−i,i4=1i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1

z=a+bi  ⟹  zˉ=a−biz = a + bi \implies \bar{z} = a - bi

(a+bi)(c+di)=(ac−bd)+(ad+bc)i(a + bi)(c + di) = (ac - bd) + (ad + bc)i

zzˉ=(a+bi)(a−bi)=a2+b2z \bar{z} = (a + bi)(a - bi) = a^2 + b^2

a+bic+di=(a+bi)(c−di)c2+d2\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{c^2 + d^2}

💡Examples

Problem 1:

Simplify the expression (3+2i)(1−4i)(3 + 2i)(1 - 4i) and write the result in the form a+bia + bi.

Solution:

(3+2i)(1−4i)=3(1)+3(−4i)+2i(1)+2i(−4i)(3 + 2i)(1 - 4i) = 3(1) + 3(-4i) + 2i(1) + 2i(-4i) =3−12i+2i−8i2= 3 - 12i + 2i - 8i^2 =3−10i−8(−1)= 3 - 10i - 8(-1) =3−10i+8= 3 - 10i + 8 =11−10i= 11 - 10i

Explanation:

Expand the brackets using the distributive law. Substitute −1-1 for i2i^2, then group the real terms and imaginary terms.

Problem 2:

Express 5+i2−3i\frac{5 + i}{2 - 3i} in the form a+bia + bi.

Solution:

5+i2−3i=(5+i)(2+3i)(2−3i)(2+3i)\frac{5 + i}{2 - 3i} = \frac{(5 + i)(2 + 3i)}{(2 - 3i)(2 + 3i)} =10+15i+2i+3i222+32= \frac{10 + 15i + 2i + 3i^2}{2^2 + 3^2} =10+17i−34+9= \frac{10 + 17i - 3}{4 + 9} =7+17i13=713+1713i= \frac{7 + 17i}{13} = \frac{7}{13} + \frac{17}{13}i

Explanation:

To divide, multiply the numerator and denominator by the conjugate of the denominator, which is 2+3i2 + 3i. This turns the denominator into a real number a2+b2a^2 + b^2.

Problem 3:

Find the real values of xx and yy such that (x+iy)(2−i)=5+5i(x + iy)(2 - i) = 5 + 5i.

Solution:

2x−xi+2yi−yi2=5+5i2x - xi + 2yi - yi^2 = 5 + 5i (2x+y)+(2y−x)i=5+5i(2x + y) + (2y - x)i = 5 + 5i Equating real and imaginary parts: 2x+y=5(Equation 1)2x + y = 5 \quad \text{(Equation 1)} −x+2y=5(Equation 2)-x + 2y = 5 \quad \text{(Equation 2)} From (1), y=5−2xy = 5 - 2x. Substitute into (2): −x+2(5−2x)=5-x + 2(5 - 2x) = 5 −x+10−4x=5  ⟹  −5x=−5  ⟹  x=1-x + 10 - 4x = 5 \implies -5x = -5 \implies x = 1 Substituting x=1x=1 into (1): 2(1)+y=5  ⟹  y=32(1) + y = 5 \implies y = 3 So, x=1,y=3x=1, y=3.

Explanation:

Expand the left side and group into real and imaginary components. Set the real part equal to 55 and the imaginary part equal to 55 to form a system of linear equations.