12 multiple-choice questions, progressively harder.
Compute (1+i)4(1 + i)^4(1+i)4.
Solution
Correct answer: D
Square twice instead of expanding all at once. First, (1+i)2=1+2i+i2=2i(1 + i)^2 = 1 + 2i + i^2 = 2i(1+i)2=1+2i+i2=2i.
(1+i)4=((1+i)2)2=(2i)2=4i2(1 + i)^4 = \left((1 + i)^2\right)^2 = (2i)^2 = 4i^2(1+i)4=((1+i)2)2=(2i)2=4i2
Since i2=−1i^2 = -1i2=−1, the answer is −4-4−4.
Which complex number satisfies z2=2iz^2 = 2iz2=2i?
Correct answer: A
Test the candidates by squaring.
(1+i)2=1+2i+i2=2i(1 + i)^2 = 1 + 2i + i^2 = 2i(1+i)2=1+2i+i2=2i
So 1+i1 + i1+i works. The others do not: (1−i)2=−2i(1 - i)^2 = -2i(1−i)2=−2i, (2i)2=−4(2i)^2 = -4(2i)2=−4, and i2=−1i^2 = -1i2=−1. (The opposite, −1−i-1 - i−1−i, would also square to 2i2i2i, but it is not among the choices.)
Let z=5−2iz = 5 - 2iz=5−2i and w=1+3iw = 1 + 3iw=1+3i. Compute z w‾z\,\overline{w}zw.
First conjugate www: w‾=1−3i\overline{w} = 1 - 3iw=1−3i. Then multiply.
(5−2i)(1−3i)=5−15i−2i+6i2=5−17i−6(5 - 2i)(1 - 3i) = 5 - 15i - 2i + 6i^2 = 5 - 17i - 6(5−2i)(1−3i)=5−15i−2i+6i2=5−17i−6
The 6i26i^26i2 becomes −6-6−6, so the real part is 5−6=−15 - 6 = -15−6=−1.
z w‾=−1−17iz\,\overline{w} = -1 - 17izw=−1−17i
Find the real numbers xxx and yyy with (x+yi)(1−i)=6−2i(x + yi)(1 - i) = 6 - 2i(x+yi)(1−i)=6−2i.
Correct answer: B
Divide both sides by 1−i1 - i1−i using the conjugate 1+i1 + i1+i; the denominator becomes 12+12=21^2 + 1^2 = 212+12=2.
x+yi=(6−2i)(1+i)2=6+6i−2i−2i22=8+4i2=4+2ix + yi = \frac{(6 - 2i)(1 + i)}{2} = \frac{6 + 6i - 2i - 2i^2}{2} = \frac{8 + 4i}{2} = 4 + 2ix+yi=2(6−2i)(1+i)=26+6i−2i−2i2=28+4i=4+2i
Matching parts gives x=4x = 4x=4 and y=2y = 2y=2. Check: (4+2i)(1−i)=4−4i+2i−2i2=6−2i(4 + 2i)(1 - i) = 4 - 4i + 2i - 2i^2 = 6 - 2i(4+2i)(1−i)=4−4i+2i−2i2=6−2i.
Compute 1+2i1−2i\dfrac{1 + 2i}{1 - 2i}1−2i1+2i.
Correct answer: C
Multiply the top and bottom by the conjugate of the denominator, 1+2i1 + 2i1+2i; the denominator becomes 12+22=51^2 + 2^2 = 512+22=5.
(1+2i)25=1+4i+4i25=−3+4i5\frac{(1 + 2i)^2}{5} = \frac{1 + 4i + 4i^2}{5} = \frac{-3 + 4i}{5}5(1+2i)2=51+4i+4i2=5−3+4i
Splitting over the denominator gives −35+45i-\dfrac{3}{5} + \dfrac{4}{5}i−53+54i.
For which real values of kkk does (2+ki)(2−ki)=13(2 + ki)(2 - ki) = 13(2+ki)(2−ki)=13?
The left side is a conjugate product, so it equals 22+k22^2 + k^222+k2.
4+k2=13 ⇒ k2=94 + k^2 = 13 \ \Rightarrow \ k^2 = 94+k2=13 ⇒ k2=9
A real number squaring to 999 can be 333 or −3-3−3, and both check, so the answer is k=3k = 3k=3 or k=−3k = -3k=−3.
What is the reciprocal of 3−4i3 - 4i3−4i, written in the form a+bia + bia+bi?
Use the reciprocal formula: multiply the top and bottom of 13−4i\frac{1}{3 - 4i}3−4i1 by the conjugate 3+4i3 + 4i3+4i.
13−4i=3+4i32+42=3+4i25=325+425 i\frac{1}{3 - 4i} = \frac{3 + 4i}{3^2 + 4^2} = \frac{3 + 4i}{25} = \frac{3}{25} + \frac{4}{25}\,i3−4i1=32+423+4i=253+4i=253+254i
Check: (3−4i)(3+4i25)=2525=1(3 - 4i)\left(\frac{3 + 4i}{25}\right) = \frac{25}{25} = 1(3−4i)(253+4i)=2525=1.
Compute 1+i1−i+1−i1+i\dfrac{1 + i}{1 - i} + \dfrac{1 - i}{1 + i}1−i1+i+1+i1−i.
Rationalize each fraction with its denominator's conjugate; both denominators become 222.
(1+i)22=2i2=i,(1−i)22=−2i2=−i\frac{(1 + i)^2}{2} = \frac{2i}{2} = i, \qquad \frac{(1 - i)^2}{2} = \frac{-2i}{2} = -i2(1+i)2=22i=i,2(1−i)2=2−2i=−i
The two results are opposites, so the sum is i+(−i)=0i + (-i) = 0i+(−i)=0.
For z=2−iz = 2 - iz=2−i, compute z2−4z+5z^2 - 4z + 5z2−4z+5.
Compute the square first: z2=(2−i)2=4−4i+i2=3−4iz^2 = (2 - i)^2 = 4 - 4i + i^2 = 3 - 4iz2=(2−i)2=4−4i+i2=3−4i.
Then assemble the expression, distributing the −4-4−4.
(3−4i)−(8−4i)+5=(3−8+5)+(−4+4) i=0(3 - 4i) - (8 - 4i) + 5 = (3 - 8 + 5) + (-4 + 4)\,i = 0(3−4i)−(8−4i)+5=(3−8+5)+(−4+4)i=0
So 2−i2 - i2−i is a root of z2−4z+5=0z^2 - 4z + 5 = 0z2−4z+5=0, the conjugate of the root 2+i2 + i2+i.
Compute (12+12i)(1−i)\left(\dfrac{1}{2} + \dfrac{1}{2}i\right)(1 - i)(21+21i)(1−i).
Expand the product and spend i2=−1i^2 = -1i2=−1.
12−12i+12i−12i2=12+12=1\frac{1}{2} - \frac{1}{2}i + \frac{1}{2}i - \frac{1}{2}i^2 = \frac{1}{2} + \frac{1}{2} = 121−21i+21i−21i2=21+21=1
The two numbers are reciprocals of each other, which is exactly what the conjugate division formula predicts.
For z=1+iz = 1 + iz=1+i, compute z3−z2−zz^3 - z^2 - zz3−z2−z.
Build the powers step by step: z2=(1+i)2=2iz^2 = (1 + i)^2 = 2iz2=(1+i)2=2i, and
z3=z2⋅z=2i(1+i)=2i+2i2=−2+2iz^3 = z^2 \cdot z = 2i(1 + i) = 2i + 2i^2 = -2 + 2iz3=z2⋅z=2i(1+i)=2i+2i2=−2+2i
Now combine all three terms.
(−2+2i)−2i−(1+i)=(−2−1)+(2−2−1) i=−3−i(-2 + 2i) - 2i - (1 + i) = (-2 - 1) + (2 - 2 - 1)\,i = -3 - i(−2+2i)−2i−(1+i)=(−2−1)+(2−2−1)i=−3−i
Two complex numbers uuu and vvv satisfy u⋅v=0u \cdot v = 0u⋅v=0. What must be true?
Suppose u≠0u \ne 0u=0. Then uuu has a reciprocal, because every nonzero complex number does.
v=1⋅v=(1u⋅u)v=1u (uv)=1u⋅0=0v = 1 \cdot v = \left(\frac{1}{u} \cdot u\right) v = \frac{1}{u}\,(u v) = \frac{1}{u} \cdot 0 = 0v=1⋅v=(u1⋅u)v=u1(uv)=u1⋅0=0
So if either factor is nonzero, the other must be 000: at least one of them is 000. Both need not be, since 1⋅0=01 \cdot 0 = 01⋅0=0.
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