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Real and Complex Numbers: Algebra 2 Math Worksheet Answer Key – PDF Download

    This algebra 2 real worksheet answer key solves all 24 problems of the real and complex numbers worksheet step by step.

    Every problem from the worksheet is solved step by step below. Final answers are in bold, and graph problems show the finished picture. Check each answer against your own work and look for the typical mistake: forgetting to replace i² with −1.

    Algebra 2Common Core: N-RN.A.1 · N-RN.A.2 · N-CN.A.1 · N-CN.A.2

    Algebra 2 Real Worksheet Answer Key: problems 1–5

    Problem 1: Classify numbers

    Natural numbers start at 1. A number is rational if it can be written as a quotient of integers.

    a) natural, integer, rational, real b) integer, rational, real c) rational, real d) irrational, real e) integer, rational, real (0 is not natural)

    Problem 2: True or false

    a) An integer \(n\) equals \(\frac{n}{1}\). b) The real numbers include all irrational numbers. c) \(\sqrt{49}=7\), which is rational. d) \(0.125=\frac{1}{8}\). e) \(\sqrt{2}\) is real but not rational.

    a) True b) True c) False d) False e) False

    Problem 3: Match points on a number line

    Estimate: \(\sqrt{5}\approx 2.24\) and \(\sqrt{7}\approx 2.65\). The order from left to right is \(-2.24\), \(-0.5\), \(1.5\), \(2.65\).

    A = \(-\sqrt{5}\), B = \(-\frac{1}{2}\), C = \(\frac{3}{2}\), D = \(\sqrt{7}\). The rational numbers are \(-\frac{1}{2}\) and \(\frac{3}{2}\).

    Problem 4: Regions of the number diagram

    The innermost region holds the natural numbers, the next the integers, the next the rational numbers. The region outside the rational numbers, still inside the real numbers, holds the irrational numbers.

    \(12\) is natural; \(-5\) is an integer but not natural; \(\frac{2}{7}\) is rational but not an integer; \(\sqrt{11}\) is irrational.

    A natural, B integers, C rational, D irrational. \(12\) is in A, \(-5\) in B, \(\frac{2}{7}\) in C, \(\sqrt{11}\) in D.

    Real numbersC Rational2/7B Integers−5A Natural12D Irrational√11
    The four numbers placed in their smallest regions.

    Problem 5: Square roots of negative numbers

    Use \(\sqrt{-a}=i\sqrt{a}\).

    a) \(\sqrt{-49}=7i\). b) \(\sqrt{-18}=i\sqrt{9\cdot 2}=3i\sqrt{2}\). c) \(\sqrt{-75}=i\sqrt{25\cdot 3}=5i\sqrt{3}\). d) \(-\sqrt{-81}=-9i\).

    a) \(7i\) b) \(3i\sqrt{2}\) c) \(5i\sqrt{3}\) d) \(-9i\)

    Algebra 2 Real Worksheet Answer Key: problems 6–10

    Problem 6: Equations with imaginary solutions

    a) \(x=\pm\sqrt{-64}=\pm 8i\). b) \(x^{2}=-45\), so \(x=\pm i\sqrt{45}=\pm 3i\sqrt{5}\). c) \(2x^{2}=-50\), \(x^{2}=-25\), \(x=\pm 5i\).

    a) \(x=\pm 8i\) b) \(x=\pm 3i\sqrt{5}\) c) \(x=\pm 5i\)

    Problem 7: Real and imaginary parts

    Write each number as \(a+bi\). The imaginary part is the real number \(b\), without the \(i\).

    a) 6 and 9 b) \(-4\) and \(-7\) c) 0 and 12 d) 15 and 0 e) \(\frac{1}{2}\) and \(-1\)

    Problem 8: Read points in the complex plane

    Read the real part across the horizontal axis, then the imaginary part up or down.

    P = \(4+2i\), Q = \(-3+i\), R = \(-2-4i\), S = \(3-3i\)

    ReIm−5−4−3−2−112345−5i−4i−3i−2i−ii2i3i4i5i4 + 2i−3 + i−2 − 4i3 − 3i
    Reading across first, then up: P = 4 + 2i, Q = −3 + i, R = −2 − 4i, S = 3 − 3i.

    Problem 9: Plot complex numbers

    Move across the real axis by the real part and up or down by the imaginary part. A is 3 right and 2 up, B is 4 left and 1 up, C is 2 left and 3 down, D is 5 right and 1 down.

    The points are plotted as shown.

    ReIm−5−4−3−2−112345−4i−3i−2i−ii2i3i4iABCD
    A = 3 + 2i, B = −4 + i, C = −2 − 3i, D = 5 − i.

    Problem 10: Add complex numbers

    a) \(3+2=5\) and \(5i-8i=-3i\). b) \(-4+6=2\) and \(i+9i=10i\). c) \(7-7=0\) and \(-2i+2i=0\).

    a) \(5-3i\) b) \(2+10i\) c) \(0\)

    Algebra 2 Real Worksheet Answer Key: problems 11–15

    Problem 11: Subtract complex numbers

    Distribute the minus sign to both parts of the second number.

    a) \(9-3=6\), \(4i-6i=-2i\). b) \(2+3=5\), \(-5i-i=-6i\). c) \(-1+1=0\), \(8i-3i=5i\).

    a) \(6-2i\) b) \(5-6i\) c) \(5i\)

    Problem 12: Add arrows in the plane

    The arrow \(z_1\) ends at 4 across and 1 up. The arrow \(z_2\) ends at 1 left and 3 up.

    Sum: \((4+i)+(-1+3i)=3+4i\). Difference: \((4+i)-(-1+3i)=5-2i\).

    \(z_1=4+i\), \(z_2=-1+3i\), \(z_1+z_2=3+4i\), \(z_1-z_2=5-2i\)

    ReIm−3−2−1123456−ii2i3i4i5i6isumz₁z₂
    The sum is the diagonal of the parallelogram.

    Problem 13: Multiply by an imaginary number

    a) \(6i+15i^{2}=6i-15\). b) \(-8i+6i^{2}=-8i-6\). c) \(7i-i^{2}=7i+1\).

    a) \(-15+6i\) b) \(-6-8i\) c) \(1+7i\)

    Problem 14: Multiply two binomials

    a) \(3+i+6i+2i^{2}=3+7i-2\). b) \(5+20i-2i-8i^{2}=5+18i+8\). c) \(9-3i-3i+i^{2}=9-6i-1\).

    a) \(1+7i\) b) \(13+18i\) c) \(8-6i\)

    Problem 15: Multiply conjugates

    The middle terms cancel, leaving the real number \(a^{2}+b^{2}\).

    a) \(16-9i^{2}=16+9=25\). b) \(36-i^{2}=37\). c) \(1-3i^{2}=1+3=4\).

    a) 25 b) 37 c) 4

    Algebra 2 Real Worksheet Answer Key: problems 16–20

    Problem 16: Powers of i

    Divide each exponent by 4 and use the remainder.

    1. 6 = 4(1) + 2, so \(i^{6}=i^{2}=-1\).
    2. 13 = 4(3) + 1, so \(i^{13}=i\).
    3. 30 = 4(7) + 2, so \(i^{30}=-1\).
    4. 47 = 4(11) + 3, so \(i^{47}=-i\).
    5. 64 = 4(16) + 0, so \(i^{64}=1\).

    a) \(-1\) b) \(i\) c) \(-1\) d) \(-i\) e) \(1\)

    × i× i× i× i1i−1−icycle of 4
    Remainder 0 gives 1, 1 gives i, 2 gives −1, 3 gives −i.

    Problem 17: Simplify with powers of i

    a) \(2(-i)+5(-1)-1+6=-2i-5-1+6\), which is \(-2i\). b) \(9i^{2}=-9\). c) \((-2)^{3}i^{3}=-8(-i)=8i\).

    a) \(-2i\) b) \(-9\) c) \(8i\)

    Problem 18: Verify a solution

    Substitute: \((1+2i)^{2}=1+4i+4i^{2}=-3+4i\). Then \(-2(1+2i)=-2-4i\).

    Add: \((-3+4i)+(-2-4i)+5=0\).

    The left side equals 0, so \(1+2i\) is a solution.

    Problem 19: Impedance in a circuit

    Add real parts: \(3+2=5\). Add imaginary parts: \(4i-i=3i\).

    \(5+3i\) ohms

    Problem 20: Voltage across a circuit

    \((2+i)(3-2i)=6-4i+3i-2i^{2}=6-i+2\).

    \(8-i\) volts

    Algebra 2 Real Worksheet Answer Key: problems 21–24

    Problem 21: Find the error

    The rule \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\) does not work when both numbers under the roots are negative. Noah must rewrite each root with \(i\) first.

    \(\sqrt{-4}=2i\) and \(\sqrt{-9}=3i\), so \(2i\cdot 3i=6i^{2}=-6\).

    The correct value is \(-6\).

    Problem 22: Moving on a game map

    After the first move: \((3+2i)+(-5+4i)=-2+6i\). After the second move: \((-2+6i)+(2-7i)=0-i\).

    The character ends at \(-i\), which is the point \((0,-1)\).

    Problem 23: Rotating a drone

    One time: \(i(4+3i)=4i+3i^{2}=-3+4i\). Two times : \(i(-3+4i)=-3i+4i^{2}=-4-3i\). Three times : \(i(-4-3i)=-4i-3i^{2}=3-4i\).

    \(-3+4i\), \(-4-3i\), \(3-4i\). Each multiplication by \(i\) turns the point a quarter turn (\(90^{\circ}\)) counterclockwise about the origin.

    ReIm−5−4−3−2−112345−5i−4i−3i−2i−ii2i3i4i5i4 + 3i−3 + 4i−4 − 3i3 − 4i
    Each multiplication by i turns the point a quarter turn counterclockwise about the origin.

    Problem 24: Make a product real

    Expand: \(2k-ki+6i-3i^{2}=(2k+3)+(6-k)i\). The product is real when the imaginary part is zero: \(6-k=0\), so \(k=6\).

    Then the real part is \(2(6)+3=15\).

    \(k=6\), and the product is 15.

    How to use this algebra 2 real worksheet answer key

    Read this algebra 2 real worksheet answer key after trying each problem yourself, and redo any problem you missed.

    Next step after this algebra 2 real worksheet answer key: Real and Complex Numbers test

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    Algebra 2 Real and Complex Numbers worksheet answer key

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