12 multiple-choice questions, progressively harder.
Write 7+i2−3i\dfrac{7 + i}{2 - 3i}2−3i7+i in standard form.
Solution
Correct answer: A
Multiply the top and bottom by the conjugate 2+3i2 + 3i2+3i, then use i2=−1i^2 = -1i2=−1.
7+i2−3i=(7+i)(2+3i)(2−3i)(2+3i)=14+21i+2i+3i24−9i2=11+23i13=1113+2313i\frac{7 + i}{2 - 3i} = \frac{(7 + i)(2 + 3i)}{(2 - 3i)(2 + 3i)} = \frac{14 + 21i + 2i + 3i^2}{4 - 9i^2} = \frac{11 + 23i}{13} = \frac{11}{13} + \frac{23}{13}i2−3i7+i=(2−3i)(2+3i)(7+i)(2+3i)=4−9i214+21i+2i+3i2=1311+23i=1311+1323i
Since 3i2=−33i^2 = -33i2=−3 and −9i2=9-9i^2 = 9−9i2=9, the fraction is 11+23i13\dfrac{11 + 23i}{13}1311+23i, split into standard form.
Simplify (1−i)3(1 - i)^3(1−i)3.
Correct answer: B
Square first, then multiply by one more factor.
(1−i)2=−2i,(−2i)(1−i)=−2i+2i2=−2−2i(1 - i)^2 = -2i, \qquad (-2i)(1 - i) = -2i + 2i^2 = -2 - 2i(1−i)2=−2i,(−2i)(1−i)=−2i+2i2=−2−2i
Here (1−i)2=1−2i+i2=−2i(1 - i)^2 = 1 - 2i + i^2 = -2i(1−i)2=1−2i+i2=−2i, and multiplying by 1−i1 - i1−i gives −2−2i-2 - 2i−2−2i.
If z=4−3iz = 4 - 3iz=4−3i, find zz‾z\overline{z}zz.
Correct answer: D
The product of a number and its conjugate is a2+b2a^2 + b^2a2+b2.
zz‾=(4−3i)(4+3i)=42+32=16+9=25z\overline{z} = (4 - 3i)(4 + 3i) = 4^2 + 3^2 = 16 + 9 = 25zz=(4−3i)(4+3i)=42+32=16+9=25
The result is real and nonnegative; getting 16−9=716 - 9 = 716−9=7 forgets that i2=−1i^2 = -1i2=−1.
Write 10i3+i\dfrac{10i}{3 + i}3+i10i in standard form.
Correct answer: C
Multiply the top and bottom by the conjugate 3−i3 - i3−i. The denominator becomes 32+12=103^2 + 1^2 = 1032+12=10.
10i3+i=10i(3−i)10=10+30i10=1+3i\frac{10i}{3 + i} = \frac{10i(3 - i)}{10} = \frac{10 + 30i}{10} = 1 + 3i3+i10i=1010i(3−i)=1010+30i=1+3i
The numerator is 30i−10i2=10+30i30i - 10i^2 = 10 + 30i30i−10i2=10+30i, then each part is divided by 101010.
Compute (2+i)2(2−i)2(2 + i)^2 (2 - i)^2(2+i)2(2−i)2.
Group the factors as a conjugate pair first: (2+i)(2−i)=5(2 + i)(2 - i) = 5(2+i)(2−i)=5.
(2+i)2(2−i)2=[(2+i)(2−i)]2=52=25(2 + i)^2 (2 - i)^2 = \left[(2 + i)(2 - i)\right]^2 = 5^2 = 25(2+i)2(2−i)2=[(2+i)(2−i)]2=52=25
Pairing the conjugates turns the whole product into 52=255^2 = 2552=25.
For which real number xxx is (x+2i)(3−i)(x + 2i)(3 - i)(x+2i)(3−i) a real number?
Expand and set the imaginary part to zero.
(x+2i)(3−i)=(3x+2)+(6−x)i,6−x=0 ⟹ x=6(x + 2i)(3 - i) = (3x + 2) + (6 - x)i, \qquad 6 - x = 0 \implies x = 6(x+2i)(3−i)=(3x+2)+(6−x)i,6−x=0⟹x=6
The product is real exactly when its imaginary part 6−x6 - x6−x vanishes, so x=6x = 6x=6.
Simplify (3+i)(3−i)(1+2i)(3 + i)(3 - i)(1 + 2i)(3+i)(3−i)(1+2i).
Multiply the conjugate pair first: (3+i)(3−i)=9+1=10(3 + i)(3 - i) = 9 + 1 = 10(3+i)(3−i)=9+1=10.
(3+i)(3−i)(1+2i)=10(1+2i)=10+20i(3 + i)(3 - i)(1 + 2i) = 10(1 + 2i) = 10 + 20i(3+i)(3−i)(1+2i)=10(1+2i)=10+20i
The conjugate pair gives the real number 101010, which then scales both parts of 1+2i1 + 2i1+2i.
Write 1+i1−i\dfrac{1 + i}{1 - i}1−i1+i 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.
1+i1−i=(1+i)22=2i2=i\frac{1 + i}{1 - i} = \frac{(1 + i)^2}{2} = \frac{2i}{2} = i1−i1+i=2(1+i)2=22i=i
The numerator (1+i)2=2i(1 + i)^2 = 2i(1+i)2=2i, then dividing by 222 leaves iii.
After simplifying 3−4i5\dfrac{3 - 4i}{5}53−4i, what is its imaginary part?
Divide each part by 555 to reach standard form.
3−4i5=35−45i\frac{3 - 4i}{5} = \frac{3}{5} - \frac{4}{5}i53−4i=53−54i
The imaginary part is the coefficient −45-\frac{4}{5}−54, not −4-4−4 (undivided) and not −45i-\frac{4}{5}i−54i.
Simplify (5+2i)(5−2i)+(1+i)2(5 + 2i)(5 - 2i) + (1 + i)^2(5+2i)(5−2i)+(1+i)2.
The first product is a conjugate pair; the second is a square.
(5+2i)(5−2i)+(1+i)2=(25+4)+2i=29+2i(5 + 2i)(5 - 2i) + (1 + i)^2 = (25 + 4) + 2i = 29 + 2i(5+2i)(5−2i)+(1+i)2=(25+4)+2i=29+2i
Here (5+2i)(5−2i)=29(5 + 2i)(5 - 2i) = 29(5+2i)(5−2i)=29 and (1+i)2=2i(1 + i)^2 = 2i(1+i)2=2i.
If (3+2i)+(x+yi)=7−i(3 + 2i) + (x + yi) = 7 - i(3+2i)+(x+yi)=7−i for real numbers xxx and yyy, what is x+yx + yx+y?
Two complex numbers are equal only when their real and imaginary parts match separately.
3+x=7 ⟹ x=4,2+y=−1 ⟹ y=−33 + x = 7 \implies x = 4, \qquad 2 + y = -1 \implies y = -33+x=7⟹x=4,2+y=−1⟹y=−3
So x+y=4+(−3)=1x + y = 4 + (-3) = 1x+y=4+(−3)=1.
Write (1+2i)22−i\dfrac{(1 + 2i)^2}{2 - i}2−i(1+2i)2 in standard form.
First (1+2i)2=−3+4i(1 + 2i)^2 = -3 + 4i(1+2i)2=−3+4i, then divide by 2−i2 - i2−i using the conjugate 2+i2 + i2+i, with denominator 22+12=52^2 + 1^2 = 522+12=5.
−3+4i2−i=(−3+4i)(2+i)5=−10+5i5=−2+i\frac{-3 + 4i}{2 - i} = \frac{(-3 + 4i)(2 + i)}{5} = \frac{-10 + 5i}{5} = -2 + i2−i−3+4i=5(−3+4i)(2+i)=5−10+5i=−2+i
The numerator is −6−3i+8i+4i2=−10+5i-6 - 3i + 8i + 4i^2 = -10 + 5i−6−3i+8i+4i2=−10+5i, then each part is divided by 555.
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