Download Assignment 7 MATH 1200 SOLUTION (1) Problem 4 from Exercises

Assignment 7
MATH 1200
SOLUTION
(1) Problem 4 from
√ Exercises for Chapter 6
(a) What is i?
SOLUTION: Writing i in polar form we have i = cos(2kπ + π2 ) + i sin(2kπ + π2 ) whenever
k ∈ Z.
So, if α denotes a square root of i then we can write α in polar coordinates as α = r(cos θ +
i sin θ). Then α2 = r2 (cos 2θ + i sin 2θ). And if α2 = i then 2θ must be of the form 2kπ + π2
and r2 = 1 so r = 1. Therefore, θ = kπ + π4 for k ∈ Z. This corresponds to two possible
√
√
√
√
values for α, namely α = 22 + 22 i (when θ = π4 ) and α = − 22 − 22 i (when θ = π + π4 )
(b) Find all the tenth roots of i. Which one is nearest to i in the Argand diagram?
SOLUTION: By the same argument as above: If α10 = i then α must be of the form
π
cos θ + i sin θ where 10θ = 2kπ + π2 . So θ is of the form 2kπ
10 + 20 for k = 1, 2, 3, .....10 (other
values of k correspond to the same complex numbers).
9π 13π 17π 21π 25π 29π 33π 37π 41π
I.e, distinct 10th roots of i have value θ = 5π
20 , 20 , 20 , 20 , 20 , 20 , 20 , 20 , 20 , 20 .
9π
π
And 20 is closest to 2 .
√
(c) Find the seven roots of the equation z 7 − 3 + i = 0. Which one of these is closest to the
imaginary axis.
√
√
11π
SOLUTION: This is equivalent to z 7 = 3−i. And 3−i = 2 cos( 11π
6 + 2kπ) + i sin( 6 + 2kπ)
for any k ∈ Z
1
Therefore, as argued above if z satisfies this equation and z = r(cos θ + i sin θ) then r = 2 7
(12k+11)π
35π 47π 59π 71π 83kπ 95π
. I.e., θ = 23π
and θ must be of the form θ = ( 11π
6 +2kπ)/7 =
42
42 , 42 , 42 , 42 , 42 , 42 , 42
11π
(the last of these has the same position as 42 ).
The imaginary axis corresponds to the angles π2 and 3π
2 . And the imaginary number cor2π
responding to position angle θ = 23π
is
within
radians
to the positive imaginary axis
42
42
π
2π
( 23π
−
=
).
And
the
next
closest
is
the
imaginary
number
corresponding to position
42
2
42
59π
4π
59π
4π
angle θ = 42 which is within 42 radians to the negative imaginary axis ( 3π
2 − 42 = 42 ).
23π
In any case the root with position angle 42 is closest to the imaginary axis.
REMARK: Alternately, on may find the roots in the above problems as described in Example 6.7. I
hadn’t yet discussed this example in class, but the general useful result is as follows:
Proposition 1. If a + bi ∈ C, then all solutions to the equation z n = a + bi can be found as follows: If
α is one solution, and w, w2 , ..., wn are all the nth roots of unity, then z = wα, w2 α, ...., wn α = α are all
solutions to the equation z n = a + bi.