12 multiple-choice questions, progressively harder.
Simplify (2+3i)2−(2−3i)2(2 + 3i)^2 - (2 - 3i)^2(2+3i)2−(2−3i)2.
Solution
Correct answer: D
Square each binomial with i2=−1i^2 = -1i2=−1, then subtract.
(2+3i)2−(2−3i)2=(−5+12i)−(−5−12i)=24i(2 + 3i)^2 - (2 - 3i)^2 = (-5 + 12i) - (-5 - 12i) = 24i(2+3i)2−(2−3i)2=(−5+12i)−(−5−12i)=24i
The real parts cancel and the imaginary parts give 12−(−12)=2412 - (-12) = 2412−(−12)=24.
Simplify i3(2+i)i^3(2 + i)i3(2+i).
Correct answer: B
Since the powers of iii cycle, i3=−ii^3 = -ii3=−i. Distribute, then replace i2i^2i2 with −1-1−1.
i3(2+i)=−i(2+i)=−2i−i2=−2i+1=1−2ii^3(2 + i) = -i(2 + i) = -2i - i^2 = -2i + 1 = 1 - 2ii3(2+i)=−i(2+i)=−2i−i2=−2i+1=1−2i
The term −i2-i^2−i2 becomes +1+1+1, the real part of the answer.
Write 14+5i3+2i\dfrac{14 + 5i}{3 + 2i}3+2i14+5i in standard form.
Correct answer: C
Multiply the top and bottom by the conjugate 3−2i3 - 2i3−2i, then use i2=−1i^2 = -1i2=−1.
14+5i3+2i=(14+5i)(3−2i)(3+2i)(3−2i)=42−28i+15i−10i29−4i2=52−13i13=4−i\frac{14 + 5i}{3 + 2i} = \frac{(14 + 5i)(3 - 2i)}{(3 + 2i)(3 - 2i)} = \frac{42 - 28i + 15i - 10i^2}{9 - 4i^2} = \frac{52 - 13i}{13} = 4 - i3+2i14+5i=(3+2i)(3−2i)(14+5i)(3−2i)=9−4i242−28i+15i−10i2=1352−13i=4−i
Since −10i2=10-10i^2 = 10−10i2=10 and −4i2=4-4i^2 = 4−4i2=4, the fraction is 52−13i13\dfrac{52 - 13i}{13}1352−13i, and dividing each part by 131313 gives 4−i4 - i4−i.
What is the conjugate of the product (3+i)(2−i)(3 + i)(2 - i)(3+i)(2−i)?
First multiply out the product, then flip the sign of the imaginary part.
(3+i)(2−i)=6−3i+2i−i2=7−i⇒7−i‾=7+i(3 + i)(2 - i) = 6 - 3i + 2i - i^2 = 7 - i \quad\Rightarrow\quad \overline{7 - i} = 7 + i(3+i)(2−i)=6−3i+2i−i2=7−i⇒7−i=7+i
The product itself is 7−i7 - i7−i; its conjugate is 7+i7 + i7+i.
Multiply (2+i)(3−2i)(1+i)(2 + i)(3 - 2i)(1 + i)(2+i)(3−2i)(1+i).
Correct answer: A
Multiply two factors first, then the third, reducing i2i^2i2 each time.
(2+i)(3−2i)=8−i,(8−i)(1+i)=9+7i(2 + i)(3 - 2i) = 8 - i, \qquad (8 - i)(1 + i) = 9 + 7i(2+i)(3−2i)=8−i,(8−i)(1+i)=9+7i
The first product is 6−4i+3i−2i2=8−i6 - 4i + 3i - 2i^2 = 8 - i6−4i+3i−2i2=8−i, and (8−i)(1+i)=8+8i−i−i2=9+7i(8 - i)(1 + i) = 8 + 8i - i - i^2 = 9 + 7i(8−i)(1+i)=8+8i−i−i2=9+7i.
Write 3+4i4+3i\dfrac{3 + 4i}{4 + 3i}4+3i3+4i in standard form.
Multiply the top and bottom by the conjugate 4−3i4 - 3i4−3i. The denominator becomes 42+32=254^2 + 3^2 = 2542+32=25.
3+4i4+3i=(3+4i)(4−3i)25=24+7i25=2425+725i\frac{3 + 4i}{4 + 3i} = \frac{(3 + 4i)(4 - 3i)}{25} = \frac{24 + 7i}{25} = \frac{24}{25} + \frac{7}{25}i4+3i3+4i=25(3+4i)(4−3i)=2524+7i=2524+257i
The numerator is 12−9i+16i−12i2=24+7i12 - 9i + 16i - 12i^2 = 24 + 7i12−9i+16i−12i2=24+7i.
Compute i+i2+i3+i4i + i^2 + i^3 + i^4i+i2+i3+i4.
Use the values of the powers of iii: i2=−1i^2 = -1i2=−1, i3=−ii^3 = -ii3=−i, i4=1i^4 = 1i4=1.
i+i2+i3+i4=i−1−i+1=0i + i^2 + i^3 + i^4 = i - 1 - i + 1 = 0i+i2+i3+i4=i−1−i+1=0
The real parts −1-1−1 and 111 cancel, and the imaginary parts iii and −i-i−i cancel.
Write 6−3i3i\dfrac{6 - 3i}{3i}3i6−3i in standard form.
The conjugate of 3i3i3i is −3i-3i−3i. Multiply top and bottom by it; the denominator becomes (3i)(−3i)=−9i2=9(3i)(-3i) = -9i^2 = 9(3i)(−3i)=−9i2=9.
6−3i3i=(6−3i)(−3i)9=−9−18i9=−1−2i\frac{6 - 3i}{3i} = \frac{(6 - 3i)(-3i)}{9} = \frac{-9 - 18i}{9} = -1 - 2i3i6−3i=9(6−3i)(−3i)=9−9−18i=−1−2i
The numerator is −18i+9i2=−9−18i-18i + 9i^2 = -9 - 18i−18i+9i2=−9−18i, then each part is divided by 999.
Which of the following equals a real number?
A number times its conjugate clears the imaginary part.
(2+5i)(2−5i)=22+52=29(2 + 5i)(2 - 5i) = 2^2 + 5^2 = 29(2+5i)(2−5i)=22+52=29
The others stay complex: (2+5i)2=−21+20i(2 + 5i)^2 = -21 + 20i(2+5i)2=−21+20i, i(2+5i)=−5+2ii(2 + 5i) = -5 + 2ii(2+5i)=−5+2i, and (2+5i)−(2−5i)=10i(2 + 5i) - (2 - 5i) = 10i(2+5i)−(2−5i)=10i.
Simplify (3+2i)2+(3−2i)2(3 + 2i)^2 + (3 - 2i)^2(3+2i)2+(3−2i)2.
Square each binomial with i2=−1i^2 = -1i2=−1, then add.
(3+2i)2+(3−2i)2=(5+12i)+(5−12i)=10(3 + 2i)^2 + (3 - 2i)^2 = (5 + 12i) + (5 - 12i) = 10(3+2i)2+(3−2i)2=(5+12i)+(5−12i)=10
The imaginary parts 12i12i12i and −12i-12i−12i cancel, leaving the real number 101010.
Simplify (2−i)3(2 - i)^3(2−i)3.
Square first, then multiply by one more factor, reducing i2i^2i2 each time.
(2−i)2=3−4i,(3−4i)(2−i)=2−11i(2 - i)^2 = 3 - 4i, \qquad (3 - 4i)(2 - i) = 2 - 11i(2−i)2=3−4i,(3−4i)(2−i)=2−11i
Here (2−i)2=4−4i+i2=3−4i(2 - i)^2 = 4 - 4i + i^2 = 3 - 4i(2−i)2=4−4i+i2=3−4i, and (3−4i)(2−i)=6−3i−8i+4i2=2−11i(3 - 4i)(2 - i) = 6 - 3i - 8i + 4i^2 = 2 - 11i(3−4i)(2−i)=6−3i−8i+4i2=2−11i.
Write 5i2−i\dfrac{5i}{2 - i}2−i5i in standard form.
Multiply the top and bottom by the conjugate 2+i2 + i2+i. The denominator becomes 22+12=52^2 + 1^2 = 522+12=5.
5i2−i=5i(2+i)5=−5+10i5=−1+2i\frac{5i}{2 - i} = \frac{5i(2 + i)}{5} = \frac{-5 + 10i}{5} = -1 + 2i2−i5i=55i(2+i)=5−5+10i=−1+2i
The numerator is 10i+5i2=−5+10i10i + 5i^2 = -5 + 10i10i+5i2=−5+10i, then each part is divided by 555.
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