Precalculus Complex Numbers and Complex Zeros Worksheet Solutions (PDF)

Precalculus math worksheet solutions: Complex Numbers and Complex Zeros – Maths PDF

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This precalculus complex numbers answers for Precalculus covers complex numbers and complex zeros (Common Core N-CN.A.1, N-CN.A.2, N-CN.A.3, N-CN.C.7, N-CN.C.8, N-CN.C.9), with a free PDF to print.

Use this key to check each complex-number problem. Every step shows where i squared becomes -1, and final answers are in bold.

Precalculus · Common Core State Standards: N-CN.A.1, N-CN.A.2, N-CN.A.3, N-CN.C.7, N-CN.C.8, N-CN.C.9

Precalculus complex numbers answers: exercises 1 to 10

Exercise 1 – Square roots of negatives

\(\sqrt{-36}=6i\). \(\sqrt{-20}=i\sqrt{20}=2i\sqrt{5}\).

\(6i\) and \(2i\sqrt{5}\)

Exercise 2 – Powers of i

\(27=4\cdot6+3\), so \(i^{27}=i^{3}=-i\). \(50=4\cdot12+2\), so \(i^{50}=i^{2}=-1\).

\(-i\) and \(-1\)

Exercise 3 – Add complex numbers

Real parts: \(4+2=6\). Imaginary parts: \(3-5=-2\).

\(6-2i\)

Exercise 4 – Subtract complex numbers

\(7-3=4\). \(-2i-(-6i)=4i\).

\(4+4i\)

Exercise 5 – Multiply

\(8-2i+12i-3i^{2}=8+10i+3\).

\(11+10i\)

Exercise 6 – Square a complex number

\(9+24i+16i^{2}=9+24i-16\).

\(-7+24i\)

Exercise 7 – Conjugate product

The conjugate is \(5+12i\). Product: \(5^{2}+12^{2}=25+144\).

\(5+12i\); \(169\)

Exercise 8 – Divide

Multiply by \(\frac{1+i}{1+i}\): numerator \(3+3i+2i+2i^{2}=1+5i\); denominator \(1+1=2\).

\(\frac12+\frac52 i\)

Exercise 9 – Divide a real number

\(\frac{10(3+i)}{(3-i)(3+i)}=\frac{30+10i}{10}\).

\(3+i\)

Exercise 10 – Modulus

\(\sqrt{25+144}=13\). \(\sqrt{9+16}=5\).

\(13\) and \(5\)

Exercises 11 and up

Exercise 11 – Read the complex plane

A = 3 + 2i, B = -2 + i, C = -1 – 3i, D = 4 – i.

Read the real part from the horizontal axis and the imaginary part from the vertical axis. \(|A|=\sqrt{9+4}=\sqrt{13}\).

\(A=3+2i,\ B=-2+i,\ C=-1-3i,\ D=4-i\); \(|A|=\sqrt{13}\)

Exercise 12 – A pure imaginary solution

\(x^{2}=-4\), so \(x=\pm\sqrt{-4}\).

\(x=\pm 2i\)

Exercise 13 – Complex roots by the formula

\(x=\frac{6\pm\sqrt{36-52}}{2}=\frac{6\pm 4i}{2}\). The discriminant is negative, so the parabola never reaches the axis.

\(x=3\pm 2i\)

Exercise 14 – Complex roots with a leading coefficient

\(x=\frac{-2\pm\sqrt{4-40}}{4}=\frac{-2\pm 6i}{4}\).

\(x=-\frac12\pm\frac32 i\)

Exercise 15 – Discriminant

Negative discriminant: no x-intercepts.

(a) \(16-16=0\): one real double root. (b) \(1-12=-11< 0\): two complex roots. (c) \(25-8=17>0\): two real roots.

(a) one real; (b) two complex; (c) two real

Exercise 16 – Conjugate pair

The conjugate \(2-3i\) is also a zero. \((x-2-3i)(x-2+3i)=(x-2)^{2}+9=x^{2}-4x+13\).

\(2-3i\); \(x^{2}-4x+13\)

Exercise 17 – All zeros of a cubic

Only one real zero.

Synthetic division by \(x-1\) gives \(x^{2}-2x+5\). Then \(x=\frac{2\pm\sqrt{4-20}}{2}=1\pm 2i\).

\(1,\ 1+2i,\ 1-2i\)

Exercise 18 – Missing zeros

Non-real zeros come in conjugate pairs: \(1-i\) and \(3i\).

\(1-i\) and \(3i\)

Exercise 19 – Impedance

Z1 + Z2 = 5 + 3i.

\(Z_{1}+Z_{2}=5+3i\), and \(\sqrt{25+9}=\sqrt{34}\approx 5.83\). \(V=(2+i)(3+4i)=6+8i+3i+4i^{2}=2+11i\).

\(5+3i\) ohms; about \(5.83\) ohms; \(V=2+11i\) volts

Exercise 20 – Verify a root

\((1+2i)^{2}=1+4i-4=-3+4i\). Then \(-3+4i-2(1+2i)+5=-3+4i-2-4i+5=0\).

The result is 0, so it is a root

Exercise 21 – Power of a complex number

\((1+i)^{2}=1+2i+i^{2}=2i\). So \((1+i)^{4}=(2i)^{2}=4i^{2}=-4\).

\(-4\)

Key facts behind this precalculus complex numbers answers

Key facts

i^2 = -1, and powers of i repeat in a cycle of 4: i, -1, -i, 1.
The conjugate of a + bi is a – bi, and (a + bi)(a – bi) = a^2 + b^2.
Modulus: |a + bi| = sqrt(a^2 + b^2), the distance from the origin in the complex plane.
A negative discriminant gives two complex roots that are conjugates.
Fundamental Theorem of Algebra: a degree-n polynomial has n complex zeros counting multiplicity.

How to use this precalculus complex numbers answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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