Download practice test 4 solutions 1.

PRACTICE TEST 4 SOLUTIONS
1.
Find the conjugate of z + w if z = 8 + 4i , w = −7 − 2i
z + w = 8 + 4i − 7 − 2i = 1 + 2i; z + w = 1 − 2i [or you could
find z and w and add them
2.
Evaluate i 59 = i 56i 3 = ( i 4 ) i 3 = i 3 = −i
3.
Write in a + bi form:
14
4.
1 + 3i
[Use rectangular form, not polar]
2−i
1 + 3i 2 + i 2 + i + 6i + 3i 2 2 − 3 + 7i
1 7
⋅
=
=
=− + i
2
2−i 2+i
4−i
5
5 5
2
Solve in the complex number system: 2 x − x + 3 = 0 [Use the quadratic formula]
5.
−b ± b 2 − 4ac − ( −1) ± ( −1) − 4 ⋅ 2 ⋅ 3 1 ± −23 1 ± 23i 1
23
x=
i
=
=
=
= ±
2a
2⋅2
4
4
4
4
Convert 3 + i to polar form. 3 + i = x + yi ; ( x, y ) = 3,1 ; r 2 = 3 + 1 = 4; r = 2 The point is in the
2
(
)
first
quadrant with cos θ =
3
1
π
, sin θ = so θ = [or 30o ]
2
2
6
3 + i = 2 ( cos 30o + i sin 30o ) or use
6.
π
6
z1
1
3
3 1
if z1 = +
i and z2 =
+ i using polar form and leave
z2
2 2
2 2
answer in a + bi form.
1
3
1
3 2 1 3
z1 = +
i ; x1 = , y1 =
, r1 = + = 1, r1 = 1; θ1 = 60o , z1 = cos 60o + i sin 60o
2 2
2
2
4 4
3 1
3
1
1 3
z1 =
+ i ; x2 =
, y2 = , r2 2 = + = 1, r2 = 1; θ 2 = 30o , z1 = cos 30o + i sin 30o
2 2
2
2
4 4
o
o
o
z1 z2 = 1 ⋅1 cos ( 30 + 60 ) + i sin ( 30 + 60o ) = cos 90o + i sin 90o = i
Find z1 z2 and
(
(
)
)
z1 / z2 = 1/1 cos ( 60o − 30o ) + i sin ( 60o − 30o ) = cos ( 30o ) + i sin ( 30o )
3 1
+ i
2 2
Find z 3 if z = 2 ( cos 40o + i sin 40o )
= cos 30o − i sin 30o =
7.
z 3 = 23 cos ( 3 ⋅ 40o ) + i sin ( 3 ⋅ 40o )  = 8 cos120o + i sin120o  = 8  − cos 60o + i sin 60o 
 1
3
= 8  − + i
 = −4 + 4i 3
2 
 2
8.
Find the 4 complex 4th roots of -16
w = −16 = 16 ( cos180o + i sin180o )
  180o 360o k 
 180o 360o k  
+
+
zk = 16 cos 
 + i sin 
  , k = 0,1, 2,3
4 
4 
 4
  4
4
zk = 2 cos ( 45o + 90o k ) + i sin ( 45o + 90o k ) 
2
2
+ 2i
= 2 + 2i
2
2

2
2
z1 = 2 cos135o + i sin135o  = 2  −
= − 2 + 2i
 + 2i
2
 2 
z0 = 2 cos 45o + 2i sin 45o = 2

2
2
z2 = 2 cos 225o + i sin 225o  = 2  −
= − 2 − 2i
 − 2i
2
 2 
9.
10.
 2
2
z3 = 2 cos 315o + i sin 315o  = 2 
= 2 − 2i
 − 2i
2
 2 
Find the focus and graph x 2 = −8 y [also find and graph the 2 points at the end of the “latus rectum”]
Note: this is a parabola symmetric in the y-axis, opening to downward, with -8 = 4(-2), so a = -2
and the focus is (0, -2). When y = - 2, x = +/- 4, so the 2 other points to graph are (4, - 2) and (-4, -2)
Find the vertex, focus and graph: ( y − 1) = 8 ( x + 1)
2
The vertex = (-1, 1); 4a = 8, a = 2; the graph is of the form Y 2 = 4aX which would be symmetric in
the x-axis, opens to the right since a > 0 so is symmetric in the line y = 1 and the focus is on this line 2
units to the right of the vertex, hence is at (1, 1). The easiest way to graph it is to graph y 2 = 8 x and
then shift it up 1 and left 1
11.
Graph the polar equations:
a)
r = 2 − 2 cos θ This is a cardiod, symmetric in the polar axis, maximum is 4 at pi, minimum at 0
and is 2 at pi/2
b)
r = 4sin θ This is a circle, symmetric in pi/2, max is 4 at pi/2, 0 at 0 [and pi]
c)
r = 3 + 2 cos θ This is a limacon, symmetric in the polar axis, max is 5 at 0, min is 1 at pi and
r = 3 at pi/2 and 3pi/2
d)
12.
a)
b)
c)
d)
r = 2sin 3θ This is a rose with 3 petals; the first petal is at sin 3theta = 1, 3theta = 90deg
theta = 30deg; the spacing is 360/3 = 120 deg; so the petals are at 30, 150, 270 deg
uuur
r r
If A = (-2, 5) and B = (4, -7) write vector AB in the form xi + yj
uuur
v
v
v
v
AB = ( 4 − ( −2 ) ) i + ( −7 − 5 ) j = 6i − 12 j
r
r
r
r
r
r
r
r
Let v = 2i + 4 j and w = −6i + 7 j . Find 3w − 4v and simplify.
r
r
r
r
r
r
r
r
3w − 4v = 3 ( −6i + 7 j ) − 4 ( 2i + 4 j ) = −26i + 5 j
r
r
r
Find a unit vector in the same direction as v = −4i + 5 j and write in
r r r
4 r
5 r
v
form xi + yj . v = 16 + 25 = 41 ; so u = −
i+
j
41
41
r r
r
r
If v = 5, write the vector v in the form ai + bj if the angle it makes with the positive x-axis is
120o . Give exact values.
 1v
v
v
3 v
5v 5 3 v
v
v = 5 ( cos120o i + sin120o j ) = 5  − i +
j  = − i +
j
2 
2
2
 2