Algebra 3 Quiz: Complex Number Operations

12 questions · exam conditions

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Complex Number OperationsQuestion 1 of 12

Evaluate $f(z)=z^2+(2-i)z+(4+3i)$ at $z=1-2i$ .

1+4i

1-6i

5-6i

-3-4i

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Algebra 3 Quiz

Algebra 3 Quiz: Complex Number Operations

Practice Complex Number Operations in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Complex Number Operations, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Evaluate $f(z)=z^2+(2-i)z+(4+3i)$ at $z=1-2i$ .

$1+4i$
$1-6i$ (correct answer)
$5-6i$
$-3-4i$

Explanation: Whenever you evaluate a complex polynomial, the core skill is using

i^2=-1

and keeping the real and imaginary parts separate. Here

z=1-2i

. Start with the quadratic term:

z^2=(1-2i)^2=1-4i+4i^2=1-4i-4=-3-4i.

Next, the linear term:

(2-i)z=(2-i)(1-2i)=2-4i-i+2i^2=0-5i.

Now add the constant

4+3i

(-3-4i)+(-5i)+(4+3i)=(-3+4)+(-4-5+3)i=1-6i.

So the value is

1-6i

. Each wrong choice reveals a different trap. The value

-3-4i

is just

z^2

; if you see it, you stopped after the first term and didn't finish evaluating the rest of the polynomial. The value

5-6i

comes from treating

i^2

+1

in the linear product, turning

0-5i

into

4-5i

; combining that with the correct

z^2

produces

5-6i

. The value

1+4i

comes from flipping the sign of the linear term, using

+5i

instead of

-5i

, so the imaginary parts combine to

+4i

instead of

-6i

. On a question like this, expand every product explicitly, then replace every

i^2

with

-1

before combining real parts and imaginary parts. And don't confuse an intermediate value with the final answer — finish all terms.

Question 2

Simplify $\frac{3-4i}{1+2i}-(1+i)$ .

$-i$
$2+3i$
$-2-3i$ (correct answer)
$-2+i$

Explanation: Whenever you see a complex fraction, the key is to create a real denominator by multiplying numerator and denominator by the conjugate of the denominator. Start with

\frac{3-4i}{1+2i}\cdot\frac{1-2i}{1-2i} =\frac{(3-4i)(1-2i)}{1+4} =\frac{-5-10i}{5} =-1-2i.

Now subtract

1+i

, being careful to distribute the minus sign to both parts:

(-1-2i)-(1+i)=-1-2i-1-i=-2-3i.

So the simplified form is

-2-3i

. The choice

-i

is a classic trap: it comes from dividing term-by-term, treating the fraction as

3/1+(-4i)/(2i)

, which is invalid because you cannot divide each term of the numerator by a single term of the denominator separately. The choice

2+3i

is simply the opposite sign of the correct answer. It results from subtracting in the wrong order: doing

(1+i)-(-1-2i)

instead of the given expression. That flips every sign. The choice

-2+i

has the correct real part but the imaginary part has the wrong sign. This often comes from using

ad-bc

instead of

bc-ad

in the quotient formula, which gives

-1+2i

for the quotient; after subtracting

1+i

, you get

-2+i

. Study tip: always rationalize withthe conjugate first, then distribute subtraction signs carefully. To check a quotient, multiply your simplified quotient back by the denominator — e.g.

(-1-2i)(1+2i)=3-4i,

so you know the first step is right.

Question 3

Simplify $(1+i)^6$ .

$8i$
$-8$
$-8i$ (correct answer)
$8$

Explanation: Whenever you see a power of a complex binomial like

(1+i)

, think in terms of magnitude and direction — or use the binomial pattern. A quick route:

(1+i)^2 = 2i

. So

(1+i)^6 = ((1+i)^2)^3 = (2i)^3 = 8i^3 = 8(-i) = -8i

. In polar form,

|1+i|=\sqrt2

and its angle is

45^\circ

, so the sixth power has magnitude

(\sqrt2)^6=8

and angle

6\cdot45^\circ=270^\circ

, which lands on the negative imaginary axis. Thus

-8i

is correct. The choice

8i

has the right magnitude but places the result on the positive imaginary axis instead of the negative one; it comes from treating

i^3

i

rather than

-i

. The choice

-8

places the result on the negative real axis, as if

i^3

became

-1

, but

i^3=-i

, not

-1

. The choice

8

also misses the imaginary direction entirely — it could come from computing only the magnitude

(\sqrt2)^6=8

or from treating

i^3

1

. Your study tip: when simplifying powers of

i

, remember the cycle

i, -1, -i, 1

. Here,

i^3

is the third term,

-i

, so

8i^3

must become

-8i

. Also, check the angle:

270^\circ

is straight down on the imaginary axis, not right or left on the real axis.

Question 4

Simplify $\frac{3-2i}{i}-4i$ .

$2-i$
$-2-i$
$-2-7i$ (correct answer)
$-2+i$

Explanation: Whenever you see an

i

in the denominator, your first move should be to remember the key fact

\frac{1}{i} = -i

, since

i(-i)=1

. That fact lets you rewrite division by

i

as multiplication by

-i

. Simplify by splitting the fraction first:

\frac{3-2i}{i}-4i=\frac{3}{i}-2-4i=3(-i)-2-4i=-2-7i.

So the result is

-2-7i

. Now look at the wrong choices. The answer

2-i

usually comes from assuming

\frac{1}{i}=i

and also distributing as if the numerator were

3+2i

, giving

3i+2-4i=2-i

. The answer

-2-i

also assumes

\frac{1}{i}=i

but keeps the subtraction:

3i-2-4i=-2-i

. These two both come from the same false idea that

\frac{1}{i}=i

. The answer

-2+i

comes from correctly simplifying to

-2-3i

, but then adding

4i

instead of subtracting it:

-2-3i+4i=-2+i

. Study tip: memorize

\frac{1}{i}=-i

, and if you ever doubt it, compute

\frac{1}{i}\cdot\frac{i}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i

. Track every negative sign carefully when distributing and subtracting; this question mainly rewards careful sign discipline.

Question 5

Let $w=(3+2i)(1-4i)$ . What is $w+\overline{w}$ ?

$22$ (correct answer)
$0$
$22-20i$
$11-10i$

Explanation: Whenever you see a complex number multiplied out and then added to its conjugate, remember the key fact:

z+\overline{z}=2\operatorname{Re}(z)

. The imaginary parts cancel, so only the real part matters. Start by multiplying:

(3+2i)(1-4i)=3-12i+2i-8i^2.

Since

i^2=-1

, the last term becomes

+8

, giving

3+8=11

for the real part and

-12i+2i=-10i

for the imaginary part. So

w=11-10i

. Its conjugate is

\overline{w}=11+10i

. Adding them:

(11-10i)+(11+10i)=22.

That is the correct answer. Now look at the wrong choices. The value

11-10i

is simply

w

itself — you forgot to add the conjugate. The value

0

would only happen if the real part of

w

were

0

, but here the real part is

11

, so the sum cannot vanish. The value

22-20i

is what you get if you add

w

to itself (doubling

w

), not

w+\overline{w}

; that mistakenly keeps the imaginary part instead of canceling it. For your study strategy: on any question involving a complex number and its conjugate, immediately think "real part doubles, imaginary part cancels." Compute

w

first, then replace every

i

with

-i

to form

\overline{w}

, and add. This avoids both the "forgot to conjugate" and "doubled the whole number" traps.

Question 6

What is the result when the product of $2+i$ and its conjugate is divided by $3-4i$ ?

$\frac{3}{5}-\frac{4}{5}i$
$3+4i$
$\frac{5}{3}-\frac{5}{4}i$
$\frac{3}{5}+\frac{4}{5}i$ (correct answer)

Explanation: Whenever you see division by a complex number, your first move should be to eliminate the imaginary part in the denominator by multiplying by its conjugate. Here, the numerator is the product of

2+i

and its conjugate

2-i

(2+i)(2-i)=4+1=5.

So the expression is

\frac{5}{3-4i}

. Multiply numerator and denominator by the conjugate of the denominator,

3+4i

\frac{5}{3-4i}\cdot\frac{3+4i}{3+4i} =\frac{15+20i}{9+16} =\frac{15+20i}{25} =\frac35+\frac45 i.

That is the correct result. The choice

\frac35-\frac45 i

is the result you would get if you mistakenly divided by

3+4i

instead of

3-4i

. The choice

3+4i

comes from forgetting to divide by the full

25

after multiplying by the conjugate — it treats the denominator as

5

instead of

25

. The choice

\frac53-\frac54 i

comes from incorrectly "splitting" the division: treating

\frac{1}{3-4i}

\frac13-\frac14 i

, which is not valid for complex numbers. Study tip: for division by

a+bi

, always multiply by

a-bi

and remember the new denominator is the real number

a^2+b^2

. That single step prevents most sign and scaling errors on this exam.

Question 7

If $(a+bi)(1+i)=3-i$ , where $a$ and $b$ are real, what is $a-b$ ?

$1$
$3$ (correct answer)
$-3$
$4$

Explanation: Whenever you see a complex number equation like

(a+bi)(1+i)=3-i

, expand the product and then match the real and imaginary parts separately. Think of

i

as a variable, but remember that

i^2=-1

. Expanding gives:

(a+bi)(1+i)=a+ai+bi+bi^2=(a-b)+(a+b)i

So the equation becomes:

(a-b)+(a+b)i=3-i

Matching real parts:

a-b=3

That is exactly what the question asks for, so the answer is

3

. If you want to confirm, matching imaginary parts gives

a+b=-1

, so

a=1

and

b=-2

, and indeed

a-b=1-(-2)=3

. Why the others are traps:

1

is the value of

a

, not

a-b

-3

b-a

, the reversed subtraction, which is a common sign error.

4

is what you might get if you incorrectly combine the coefficients, like treating

3-(-1)

as the answer instead of separating real and imaginary parts. The key takeaway: when multiplying complex numbers, expand first, then equate real parts to real parts and imaginary parts to imaginary parts. Often the quantity asked for appears directly from one of those two equations, so write them down before solving fully.

Question 8

What is the multiplicative inverse of $3-4i$ , written in the form $a+bi$ ?

$\frac{3}{25}+\frac{4}{25}i$ (correct answer)
$\frac{3}{25}-\frac{4}{25}i$
$\frac{1}{3}+\frac{1}{4}i$
$\frac{1}{3}-\frac{1}{4}i$

Explanation: When you see a question about a multiplicative inverse of a complex number, think of it as rationalizing a denominator: you want to turn the denominator into a real number. For a complex number

3-4i

, its inverse is

\frac{1}{3-4i}

. Multiply top and bottom by the conjugate,

3+4i

, to get

\frac{3+4i}{(3-4i)(3+4i)}

. The denominator simplifies to

3^2+4^2=9+16=25

, so the inverse is

\frac{3+4i}{25}=\frac{3}{25}+\frac{4}{25}i

. That matches the correct choice. Now look at the traps. The choice

\frac{3}{25}-\frac{4}{25}i

is just the original number divided by 25, or the negation of the imaginary part — it comes from forgetting that the conjugate changes the sign of the imaginary part. The choices

\frac{1}{3}+\frac{1}{4}i

and

\frac{1}{3}-\frac{1}{4}i

both incorrectly take reciprocals of the real and imaginary parts separately; but complex numbers do not work that way. You must use the entire denominator, not each term independently. A reliable strategy: whenever you need the inverse of

a+bi

, write

\frac{a-bi}{a^2+b^2}

directly. The denominator is always the real number

a^2+b^2

, and the numerator is the conjugate. Practice this pattern so you can move quickly without multiplying out every time. On the exam, watch for the answer that keeps the original sign unchanged — that is the most common trap.

Question 9

Let $z=2+i$ . What is $z^2-\overline{z}$ ?

$5+3i$
$1+5i$ (correct answer)
$1+3i$
$3+5i$

Explanation: Whenever you see a complex number like

z=2+i

raised to a power and combined with its conjugate, the key is to handle each piece separately. First, expand

z^2

using

(a+bi)^2=a^2+2abi+b^2i^2

. Here,

z^2=(2+i)^2=4+4i+i^2=4+4i-1=3+4i.

The conjugate of

2+i

\overline{z}=2-i.

Now subtract:

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

So the correct choice is

1+5i

. The other choices come from common sign mistakes. If you instead added the conjugate, you would get

(3+4i)+(2-i)=5+3i

. If you mistakenly used

2+i

as the conjugate, you would compute

(3+4i)-(2+i)=1+3i

. And if you subtracted only the imaginary part of the conjugate but forgot its real part, you'd get

3+4i-(-i)=3+5i

. A good habit: write out

z^2

and

\overline{z}

separately, replace

i^2

with

-1

immediately, then combine real parts and imaginary parts carefully. Watching the sign of the imaginary part in the conjugate is the key to avoiding most traps on this type of problem.

Question 10

Simplify $\frac{2+3i}{3-2i}$ .

$i$ (correct answer)
$\frac{12}{13}+\frac{5}{13}i$
$1$
$-i$

Explanation: When you see a division of complex numbers, your goal is to eliminate the imaginary part from the denominator. Do this by multiplying the numerator and denominator by the conjugate of the denominator. Here the denominator is

3-2i

, so its conjugate is

3+2i

. Thus:

\frac{2+3i}{3-2i}\cdot\frac{3+2i}{3+2i} =\frac{(2+3i)(3+2i)}{(3-2i)(3+2i)} =\frac{6+4i+9i+6i^2}{9-4i^2}.

Since

i^2=-1

, the numerator becomes

6+13i-6=13i

, and the denominator becomes

9+4=13

. Therefore the quotient is

\frac{13i}{13}=i

. The choice

\frac{12}{13}+\frac{5}{13}i

is the trap from multiplying the numerator by the original denominator

3-2i

instead of its conjugate; that gives

(2+3i)(3-2i)=12+5i

, but the denominator must be multiplied by the same conjugate

3+2i

to become real. The choice

1

would require

2+3i=3-2i

, which is false; you cannot cancel unrelated complex numbers. The choice

-i

reflects a sign error in computing the numerator or conjugate; the correct imaginary product is

+13i

, not

-13i

. Strategy: whenever you divide complex numbers, always multiply by the conjugate of the denominator over itself. Then check your answer by multiplying it by the original denominator:

i(3-2i)=3i-2i^2=2+3i

, which matches the original numerator.

Question 11

Simplify $i^{2025}+2i^{2026}+3i^{2027}+4i^{2028}$ .

$2+4i$
$-2-2i$
$2$
$2-2i$ (correct answer)

Explanation: Whenever you see a large power of

i

, reduce the exponent modulo

4

, because

i^4=1

and the powers repeat as

i,\,-1,\,-i,\,1

. Here,

2025\div 4

leaves remainder

1

, so

i^{2025}=i

;

2026

leaves remainder

2

, so

i^{2026}=-1

;

2027

leaves remainder

3

, so

i^{2027}=-i

; and

2028

leaves remainder

0

, so

i^{2028}=1

. Substituting gives

i+2(-1)+3(-i)+4(1)=i-2-3i+4=2-2i.

So the simplified form is

2-2i

. The choice

2+4i

usually comes from thinking

i^{2027}=i

, but

i^3=-i

, not

i

. The choice

-2-2i

often appears when students correctly reach an intermediate like

-2-2i

and then forget that multiplying by

i

changes the sign:

i(-2-2i)=-2i-2i^2=2-2i

. The choice

2

captures only the real part,

4-2=2

, while dropping the imaginary part

-2i

; if a complex number is the result, both real and imaginary parts must be kept. A good study habit: write the remainder map

0\to 1,\;1\to i,\;2\to -1,\;3\to -i

before simplifying, then combine real terms and imaginary terms separately. This makes large-exponent

i

problems fast and reliable.

Question 12

Simplify $\frac{1}{2-i}+\frac{1}{2+i}$ .

$\frac{4}{3}$
$\frac{2}{5}$
$1$
$\frac{4}{5}$ (correct answer)

Explanation: Whenever you see a sum of reciprocals involving complex conjugates, your instinct should be to combine them over a common denominator — the imaginary parts will cancel, leaving a real fraction. For

\frac{1}{2-i}+\frac{1}{2+i}

, the common denominator is

(2-i)(2+i)=4-i^2

. Since

i^2=-1

, this simplifies to

4-(-1)=5

. The numerator is

(2+i)+(2-i)=4

. So the fraction is

\frac{4}{5}

. The choice

\frac{4}{3}

comes from treating

i^2

+1

, which would make the denominator

4-1=3

. But

i^2=-1

, not

1

. The choice

\frac{2}{5}

likely comes from using

2

as the numerator — but the real parts add to

2+2=4

. The choice

1

results from ignoring the

i^2

term completely and treating the denominator product as just

4

; the conjugate product actually contributes

+1

because

-(i^2)=1

. A good memory aid: for conjugates

a+bi

and

a-bi

(a+bi)(a-bi)=a^2+b^2

. Here that denominator is

2^2+1^2=5

, and the numerator is twice the real part,

2\cdot 2=4

. Look for that pattern whenever you add reciprocals of conjugates.