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Chapter Review · a rapid pre-test review (speedrun)

Complex Numbers: Chapter Review

A rapid review before the test: the chapter's vocabulary and notation, every formula with the conditions to use it, the standard problem types step by step, and the traps that cost points.

Vocabulary and notation

Imaginary unit ii
Defined by the single equation i2=1i^2 = -1, with every ordinary law of algebra kept. No real number has a negative square, so ii lies nowhere on the real line.
Complex number, standard form a+bia + bi
Real part aa and imaginary part bb, both real numbers. The imaginary part is the coefficient bb with its sign, never bibi: for 74i7 - 4i it is 4-4.
Pure imaginary number
A number bibi with b0b \neq 0, so its real part is 00. Taking b=0b = 0 instead gives a real number: the reals sit inside the complex numbers unchanged.
Conjugate z\overline{z}
For z=a+biz = a + bi, the number abia - bi: only the sign of the imaginary part flips. The conjugate of 35i-3 - 5i is 3+5i-3 + 5i, not 3+5i3 + 5i.
Conjugate pair
p+qip + qi and pqip - qi with q0q \neq 0: equal real parts, opposite imaginary parts, mirror images across the real axis.
Complex plane (Argand diagram)
The plane in which a+bia + bi is the point (a,b)(a, b). The horizontal real axis holds the numbers with b=0b = 0; the vertical imaginary axis holds 00 and the pure imaginaries.
Discriminant Δ=b24ac\Delta = b^2 - 4ac
For REAL a0a \neq 0, bb, cc it no longer counts roots (there are always two with multiplicity); it locates them: Δ>0\Delta > 0 two real, Δ=0\Delta = 0 one repeated real, Δ<0\Delta < 0 a non-real conjugate pair. Let one coefficient go non-real and none of it holds: x22ix1x^2 - 2ix - 1 has Δ=0\Delta = 0 and the repeated root ii.

Formulas and theorems

  • The definition of ii, and roots of negatives

    i2=1,k=iki^2 = -1, \qquad \sqrt{-k} = i\sqrt{k}

    Use when k>0k > 0. The symbol k\sqrt{-k} names ONE number, iki\sqrt{k}, while x2=kx^2 = -k has two solutions, ±ik\pm i\sqrt{k}. Keep the ii outside the radical.

  • Powers of ii

    i1=i,i2=1,i3=i,i4=1in=ir  for  n=4q+r\begin{gathered} i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1 \\ i^n = i^r \ \text{ for } \ n = 4q + r \end{gathered}

    Use when nn a positive integer, rr its remainder on division by 44; remainder 00 gives 11. Parity decides nothing: i6=1i^6 = -1 but i8=1i^8 = 1. Any four consecutive powers sum to 00.

  • Equality of complex numbers

    a+bi=c+di    a=c  and  b=da + bi = c + di \iff a = c \ \text{ and } \ b = d

    Use when aa, bb, cc, dd ALL real. One complex equation therefore carries two real ones. The complex numbers have no order, so << and >> between them are meaningless.

  • Addition and subtraction

    (a+bi)±(c+di)=(a±c)+(b±d)i(a + bi) \pm (c + di) = (a \pm c) + (b \pm d)\,i

    Use when Any two complex numbers. Distribute a minus sign through BOTH terms of the second. Geometrically, adding c+dic + di slides every point cc across and dd up; zwz - w is the arrow from ww to zz.

  • Multiplication

    (a+bi)(c+di)=(acbd)+(ad+bc)i\begin{gathered} (a + bi)(c + di) \\ = (ac - bd) + (ad + bc)\,i \end{gathered}

    Use when Any two complex numbers. Rebuild it by expanding rather than memorizing: each i2i^2 becomes 1-1 and joins the real part with its sign flipped.

  • Conjugate product

    zz=(a+bi)(abi)=a2+b2=z2z\,\overline{z} = (a + bi)(a - bi) = a^2 + b^2 = |z|^2

    Use when Any z=a+biz = a + bi. The value is REAL, never negative, and 00 only for z=0z = 0. The i2i^2 turns the difference of squares into a SUM, which is what lets a conjugate clear a denominator.

  • Conjugation passes through arithmetic

    z+w=z+wzw=zwz=z\begin{gathered} \overline{z + w} = \overline{z} + \overline{w} \\ \overline{zw} = \overline{z}\,\overline{w} \\ \overline{\overline{z}} = z \end{gathered}
    Conjugation is the reflection across the real axisTwo marked points in the complex plane, one above the horizontal real axis and one directly below it at the same distance, joined by a dashed vertical segment. A small square marks the right angle where that segment crosses the axis, and a single tick on each half shows the two halves are equal. Faint rays from the origin to each point are the same length, so the two numbers share one modulus.ReIm0z = a + biz = a − bi
    Text description

    The point z and the point for its conjugate sit directly above and below each other at equal distances from the real axis, so conjugation reflects the plane across that axis.

    Use when Any complex zz and ww; the product rule repeats into powers. Conjugation fixes exactly the real numbers and reflects the plane across the real axis, so z=z=z|\overline{z}| = |-z| = |z|.

  • Division and reciprocals

    a+bic+di=(a+bi)(cdi)c2+d21c+di=cdic2+d2\begin{gathered} \frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{c^2 + d^2} \\ \frac{1}{c + di} = \frac{c - di}{c^2 + d^2} \end{gathered}

    Use when c+di0c + di \neq 0, the only forbidden divisor, since c2+d2=0c^2 + d^2 = 0 only for c=d=0c = d = 0. Multiply NUMERATOR AND DENOMINATOR by the DENOMINATOR's conjugate, which is multiplying by 11.

    e.g. 1i=i02+12=i\dfrac{1}{i} = \dfrac{-i}{0^2 + 1^2} = -i, and i(i)=1i(-i) = 1.

  • Modulus, and the modulus of a product

    a+bi=a2+b2,zw=zw|a + bi| = \sqrt{a^2 + b^2}, \qquad |zw| = |z|\,|w|
    The modulus is the hypotenuse of a right triangle from the originA right triangle in the complex plane with its right angle away from the origin. The horizontal leg lies along the real axis and is labelled a, the vertical leg is labelled b, and a small square marks the right angle where they meet. The highlighted hypotenuse runs from the origin to the dot for a plus b i and is labelled with the modulus, so Pythagoras gives it as the square root of a squared plus b squared.ReIm0ab|z|z = a + bi
    Text description

    The point a plus b i closes a right triangle whose legs along the axes have lengths a and b, so its hypotenuse, the distance from zero, is the modulus.

    Use when Any complex numbers. The modulus is the distance from 00: always a NONNEGATIVE REAL number, never complex, and 00 only for z=0z = 0. Multiplicativity extends to any number of factors; no such rule holds for a sum.

    e.g. (6+8i)(43i)=105=50|(6 + 8i)(4 - 3i)| = 10 \cdot 5 = 50, with no expansion.

  • Distance, and circles

    zw=(ac)2+(bd)2zz0=r\begin{gathered} |z - w| = \sqrt{(a - c)^2 + (b - d)^2} \\ |z - z_0| = r \end{gathered}

    Use when For z=a+biz = a + bi and w=c+diw = c + di; distance from 00 is the case w=0w = 0. The circle needs r>0r > 0 and has center z0z_0, with zz0<r|z - z_0| < r inside and >r> r outside.

    e.g. (1+2i)(3i)=4+3i=5|(-1 + 2i) - (3 - i)| = |-4 + 3i| = 5.

  • Triangle inequality

    zw    z+w    z+w\bigl| |z| - |w| \bigr| \;\le\; |z + w| \;\le\; |z| + |w|

    Use when Any two complex numbers. The upper bound is an equality only when the arrows point the same way, or one is 00; never solve with it as an equation.

    e.g. z=3|z| = 3 and w=4|w| = 4 pin z+w|z + w| into [1,7][1, 7].

  • Quadratic formula with a negative discriminant

    x=b±iΔ2aΔ=b24ac<0\begin{gathered} x = \frac{-b \pm i\sqrt{|\Delta|}}{2a} \\ \Delta = b^2 - 4ac < 0 \end{gathered}

    Use when aa, bb, cc real and a0a \neq 0. Divide BOTH numerator terms by 2a2a. The roots are a conjugate pair p±qip \pm qi with q0q \neq 0, sharing the real part p=b2ap = -\tfrac{b}{2a}, the axis of symmetry.

  • Conjugate root theorem

    az2+bz+c=0    az2+bz+c=0\begin{gathered} az^2 + bz + c = 0 \\ \implies a\,\overline{z}^{\,2} + b\,\overline{z} + c = 0 \end{gathered}

    Use when EVERY coefficient must be real, with a0a \neq 0. It says nothing new about a real root, where z=z\overline{z} = z; its force is that non-real roots never appear alone, so the inventories are two real, one repeated real, or one conjugate pair.

  • Sum and product of the roots, and complete factorization

    r1+r2=ba,r1r2=caax2+bx+c=a(xr1)(xr2)\begin{gathered} r_1 + r_2 = -\frac{b}{a}, \qquad r_1 r_2 = \frac{c}{a} \\ ax^2 + bx + c = a(x - r_1)(x - r_2) \end{gathered}

    Use when Any quadratic with a0a \neq 0; both relations hold whether the roots are real or not. For a non-real pair p±qip \pm qi the product is p2+q2=z2>0p^2 + q^2 = |z|^2 > 0. Over the complex numbers every quadratic splits into two linear factors.

    e.g. Roots 1±2i1 \pm 2i: x22x+5=(x12i)(x1+2i)x^2 - 2x + 5 = (x - 1 - 2i)(x - 1 + 2i).

Problem types, step by step

Simplify or combine square roots of negative numbers

  1. Rewrite every k\sqrt{-k} as iki\sqrt{k} before doing anything else with it.
  2. Simplify each real radical by pulling out perfect-square factors.
  3. Only now multiply, divide, or add, replacing every i2i^2 with 1-1.

e.g. 818=(2i2)(3i2)=6i22=12\sqrt{-8} \cdot \sqrt{-18} = \left(2i\sqrt{2}\right)\left(3i\sqrt{2}\right) = 6i^2 \cdot 2 = -12.

Evaluate a power of ii

  1. For a product or quotient of powers, combine the exponents first.
  2. Divide the exponent by 44 and keep the remainder (only the last two digits matter, since 100100 is a multiple of 44), then read the cycle: 010 \to 1, 1i1 \to i, 212 \to -1, 3i3 \to -i.
  3. For a long sum of consecutive powers, discard blocks of four, which each total 00, and evaluate the leftovers.

e.g. i45i23=i68i^{45} \cdot i^{23} = i^{68}, and 68=41768 = 4 \cdot 17 leaves remainder 00, so the value is 11.

Add, subtract, or multiply two complex numbers

  1. For a difference, distribute the minus sign through both terms of the second number first.
  2. Combine real with real and imaginary with imaginary; for a product, expand all four products as with any binomials, signs attached.
  3. Replace every i2i^2 by 1-1 and report as a+bia + bi, with no power of ii above the first.

e.g. (4+3i)(68i)=2+11i(4 + 3i) - (6 - 8i) = -2 + 11i, and (32i)2=912i+4i2=512i(3 - 2i)^2 = 9 - 12i + 4i^2 = 5 - 12i.

Divide, or solve a linear equation over the complex numbers

  1. Confirm the divisor is not 00, then multiply numerator and denominator by the conjugate of the DENOMINATOR.
  2. The denominator becomes the real c2+d2c^2 + d^2; expand the numerator, spend every i2i^2, and divide each part by it.
  3. For wz=vwz = v, compute z=v/wz = v / w this way and check by multiplying back.

e.g. 4+2i1+i=(4+2i)(1i)2=62i2=3i\dfrac{4 + 2i}{1 + i} = \dfrac{(4 + 2i)(1 - i)}{2} = \dfrac{6 - 2i}{2} = 3 - i.

Measure in the complex plane: modulus, distance, circles

  1. Plot a+bia + bi at (a,b)(a, b): real part across, imaginary part up, signs included.
  2. For z|z|, square both parts, add, then take the nonnegative square root, in that order; subtract first for a gap between two numbers; multiply the separate moduli for a product.
  3. Given zz0=r|z - z_0| = r, read the center and radius off and draw rather than solving; compare a distance to rr to place a point inside, on, or outside.
  4. To pair it with a condition on one part, fix that part in z=a+biz = a + bi and solve the real equation that results.

e.g. Modulus 55 with imaginary part 33: a2+9=25a^2 + 9 = 25, so z=4+3iz = 4 + 3i or 4+3i-4 + 3i.

Solve a quadratic whose discriminant is negative

  1. Put it as ax2+bx+c=0ax^2 + bx + c = 0 and compute Δ=b24ac\Delta = b^2 - 4ac.
  2. With Δ<0\Delta < 0, rewrite Δ\sqrt{\Delta} as iΔi\sqrt{|\Delta|}, simplify that radical, and divide BOTH numerator terms by 2a2a.
  3. Report the conjugate pair p±qip \pm qi; check pp against the axis b2a-\tfrac{b}{2a} and substitute one root back, which certifies both.

e.g. x2+2x+10=0x^2 + 2x + 10 = 0: Δ=36\Delta = -36, so x=2±6i2=1±3ix = \tfrac{-2 \pm 6i}{2} = -1 \pm 3i.

Rebuild a real quadratic from one non-real root

  1. Confirm the coefficients are required to be real, then conjugate the given root to get the second one.
  2. Add the pair for the sum 2p2p, and multiply them for the product p2+q2p^2 + q^2, the squared modulus.
  3. Write x2(sum)x+(product)x^2 - (\text{sum})\,x + (\text{product}), scaling by aa if a leading coefficient is required, then verify by solving.

e.g. Root 34i3 - 4i: sum 66, product 2525, so x26x+25=0x^2 - 6x + 25 = 0.

Split one complex equation into two real equations

  1. Simplify each side to standard form, with every part a real expression.
  2. Set the real parts equal, set the imaginary parts equal, solve the real system, and substitute back to check.

e.g. (3x1)+(y+2)i=85i(3x - 1) + (y + 2)i = 8 - 5i gives x=3x = 3 and y=7y = -7.

Exam traps

  • Trap Applying ab=ab\sqrt{a}\,\sqrt{b} = \sqrt{ab} to two negatives: 49=36=6\sqrt{-4} \cdot \sqrt{-9} = \sqrt{36} = 6.

    Fix That rule was proved for a,b0a, b \ge 0 only. Convert first: (2i)(3i)=6i2=6(2i)(3i) = 6i^2 = -6, the opposite sign.

  • Trap Leaving an i2i^2 standing, or reading it as 11.

    Fix Every i2i^2 becomes 1-1 on sight, so 810i+12i15i28 - 10i + 12i - 15i^2 finishes at 23+2i23 + 2i, not 7+2i-7 + 2i.

  • Trap Answering "no solution" when the discriminant is negative.

    Fix No REAL solution, but exactly two complex ones, a conjugate pair. Name the number system with the count.

  • Trap Dividing only part of the numerator by 2a2a, as 6±4i2=6±2i\tfrac{6 \pm 4i}{2} = 6 \pm 2i or 3±4i3 \pm 4i.

    Fix The denominator sits under the whole numerator: 6±4i2=3±2i\tfrac{6 \pm 4i}{2} = 3 \pm 2i.

  • Trap Using z2z^2 where zz=z2z\,\overline{z} = |z|^2 is meant.

    Fix (3+4i)2=7+24i(3 + 4i)^2 = -7 + 24i, while (3+4i)(34i)=25(3 + 4i)(3 - 4i) = 25. The two agree only for real zz.

  • Trap Computing the gap between zz and ww as zw|z| - |w|.

    Fix Subtract first: the distance is zw|z - w|. For z=3z = 3 and w=3iw = 3i it is 323\sqrt{2}, while zw=0|z| - |w| = 0.

  • Trap Pairing roots as conjugates without checking that the coefficients are real.

    Fix x22ix1=(xi)2x^2 - 2ix - 1 = (x - i)^2 has the double root ii, and i-i is not a root at all.

Chapter test Questions from across the chapter