12 multiple-choice questions, progressively harder.
Compute 12−i+12+i\dfrac{1}{2 - i} + \dfrac{1}{2 + i}2−i1+2+i1.
Solution
Correct answer: D
Add over the common denominator (2−i)(2+i)=22+12=5(2 - i)(2 + i) = 2^2 + 1^2 = 5(2−i)(2+i)=22+12=5.
12−i+12+i=(2+i)+(2−i)5=45\frac{1}{2 - i} + \frac{1}{2 + i} = \frac{(2 + i) + (2 - i)}{5} = \frac{4}{5}2−i1+2+i1=5(2+i)+(2−i)=54
The imaginary parts of the numerator cancel, leaving the real number 45\frac{4}{5}54.
Simplify (4+3i)(4−3i)−(2+i)(2−i)(4 + 3i)(4 - 3i) - (2 + i)(2 - i)(4+3i)(4−3i)−(2+i)(2−i).
Correct answer: C
Each product is a number times its conjugate, giving a2+b2a^2 + b^2a2+b2.
(4+3i)(4−3i)−(2+i)(2−i)=(16+9)−(4+1)=25−5=20(4 + 3i)(4 - 3i) - (2 + i)(2 - i) = (16 + 9) - (4 + 1) = 25 - 5 = 20(4+3i)(4−3i)−(2+i)(2−i)=(16+9)−(4+1)=25−5=20
Both products are real, so the difference is the real number 202020.
Simplify (5−2i)2(5 - 2i)^2(5−2i)2.
Correct answer: A
Use the square-of-a-binomial pattern, then replace i2i^2i2 with −1-1−1.
(5−2i)2=25−20i+4i2=25−20i−4=21−20i(5 - 2i)^2 = 25 - 20i + 4i^2 = 25 - 20i - 4 = 21 - 20i(5−2i)2=25−20i+4i2=25−20i−4=21−20i
The term 4i24i^24i2 becomes −4-4−4, so the real part is 25−4=2125 - 4 = 2125−4=21.
Compute 2+i1−2i+2−i1+2i\dfrac{2 + i}{1 - 2i} + \dfrac{2 - i}{1 + 2i}1−2i2+i+1+2i2−i.
Each denominator has 12+22=51^2 + 2^2 = 512+22=5 after multiplying by its conjugate.
2+i1−2i=5i5=i,2−i1+2i=−5i5=−i\frac{2 + i}{1 - 2i} = \frac{5i}{5} = i, \qquad \frac{2 - i}{1 + 2i} = \frac{-5i}{5} = -i1−2i2+i=55i=i,1+2i2−i=5−5i=−i
The two quotients are opposites, so i+(−i)=0i + (-i) = 0i+(−i)=0.
Compute (1+i)2i\dfrac{(1 + i)^2}{i}i(1+i)2.
First (1+i)2=2i(1 + i)^2 = 2i(1+i)2=2i, then divide by iii.
(1+i)2i=2ii=2\frac{(1 + i)^2}{i} = \frac{2i}{i} = 2i(1+i)2=i2i=2
The factors of iii cancel, leaving the real number 222.
Simplify i5+i10+i15i^5 + i^{10} + i^{15}i5+i10+i15.
Reduce each power using the remainder when the exponent is divided by 444.
i5+i10+i15=i+(−1)+(−i)=−1i^5 + i^{10} + i^{15} = i + (-1) + (-i) = -1i5+i10+i15=i+(−1)+(−i)=−1
Here i5=ii^5 = ii5=i, i10=i2=−1i^{10} = i^2 = -1i10=i2=−1, and i15=i3=−ii^{15} = i^3 = -ii15=i3=−i, and the iii terms cancel.
If z=2+iz = 2 + iz=2+i, what is z⋅z‾z\cdot\overline{z}z⋅z?
Correct answer: B
The product of a number and its conjugate is a2+b2a^2 + b^2a2+b2.
z⋅z‾=(2+i)(2−i)=22+12=5z\cdot\overline{z} = (2 + i)(2 - i) = 2^2 + 1^2 = 5z⋅z=(2+i)(2−i)=22+12=5
Getting 4−1=34 - 1 = 34−1=3 mistakes the sum for the real-number difference of squares.
Compute 4+2i1+i−4−2i1−i\dfrac{4 + 2i}{1 + i} - \dfrac{4 - 2i}{1 - i}1+i4+2i−1−i4−2i.
Each denominator has 12+12=21^2 + 1^2 = 212+12=2 after multiplying by its conjugate.
4+2i1+i=3−i,4−2i1−i=3+i\frac{4 + 2i}{1 + i} = 3 - i, \qquad \frac{4 - 2i}{1 - i} = 3 + i1+i4+2i=3−i,1−i4−2i=3+i
Subtracting, (3−i)−(3+i)=−2i(3 - i) - (3 + i) = -2i(3−i)−(3+i)=−2i.
Write −2+6i1+i\dfrac{-2 + 6i}{1 + i}1+i−2+6i in standard form.
Multiply the top and bottom by the conjugate 1−i1 - i1−i. The denominator becomes 12+12=21^2 + 1^2 = 212+12=2.
−2+6i1+i=(−2+6i)(1−i)2=4+8i2=2+4i\frac{-2 + 6i}{1 + i} = \frac{(-2 + 6i)(1 - i)}{2} = \frac{4 + 8i}{2} = 2 + 4i1+i−2+6i=2(−2+6i)(1−i)=24+8i=2+4i
The numerator is −2+2i+6i−6i2=4+8i-2 + 2i + 6i - 6i^2 = 4 + 8i−2+2i+6i−6i2=4+8i, then each part is divided by 222.
Simplify (1+i)2+(1−i)2(1 + i)^2 + (1 - i)^2(1+i)2+(1−i)2.
Square each binomial with i2=−1i^2 = -1i2=−1, then add.
(1+i)2+(1−i)2=2i+(−2i)=0(1 + i)^2 + (1 - i)^2 = 2i + (-2i) = 0(1+i)2+(1−i)2=2i+(−2i)=0
The two squares are 2i2i2i and −2i-2i−2i, which cancel.
Write 3+22i2+5i\dfrac{3 + 22i}{2 + 5i}2+5i3+22i in standard form.
Multiply the top and bottom by the conjugate 2−5i2 - 5i2−5i. The denominator becomes 22+52=292^2 + 5^2 = 2922+52=29.
3+22i2+5i=(3+22i)(2−5i)29=116+29i29=4+i\frac{3 + 22i}{2 + 5i} = \frac{(3 + 22i)(2 - 5i)}{29} = \frac{116 + 29i}{29} = 4 + i2+5i3+22i=29(3+22i)(2−5i)=29116+29i=4+i
The numerator is 6−15i+44i−110i2=116+29i6 - 15i + 44i - 110i^2 = 116 + 29i6−15i+44i−110i2=116+29i, then each part is divided by 292929.
For z=1+iz = 1 + iz=1+i, compute z2−2z+2z^2 - 2z + 2z2−2z+2.
Compute z2=(1+i)2=2iz^2 = (1 + i)^2 = 2iz2=(1+i)2=2i and 2z=2+2i2z = 2 + 2i2z=2+2i, then combine.
z2−2z+2=2i−(2+2i)+2=0z^2 - 2z + 2 = 2i - (2 + 2i) + 2 = 0z2−2z+2=2i−(2+2i)+2=0
Every term cancels, so the expression evaluates to 000.
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