MCS 351 ENGINEERING MATHEMATICS SOLUTION OF ...

22. Let z = 8√ 1. Since
1 = cos(2kπ) + i sin(2kπ), k = 0, ±1, ±2, ...,
Above equation can be rewrite in the following form:
There are 8 roots as following:
23. Let z = x + iy. If z = 4√ −1, then z 4 = −1.
z = 8√ 1 = cos( 2kπ
8 ) + i sin(2kπ 8 )
k = 0 =⇒ z 0 = 1,
k = 1 =⇒ z 1 = 1 √
2
+ i 1 √
2
,
k = 2 =⇒ z 2 = i,
k = 3 =⇒ z 3 = − 1 √
2
+ i 1 √
2
,
k = 4 =⇒ z 4 = −1,
k = 5 =⇒ z 5 = − 1 √
2
− i 1 √
2
,
k = 6 =⇒ z 6 = −i,
k = 7 =⇒ z 7 = 1 √
2
− i 1 √
2
,
(x+iy) 4 = (x 4 −6x 2 y 2 +y 4 )+(4x 3 y−4xy 3 )i = −1 ⇒ x 4 −6x 2 y 2 +y 4 = −1 and 4x 3 y−4xy 3 = 0.
Solving latter equation, we have x = y. After, solving first equation with x = y, we have x = y = ∓ 1 √
2
.
Therefore, z = ∓ 1 √
2
∓ 1 √
2
i.
24. Let w = 3 + 4i and z = 3√ w. From this, we get
arg(w) = α =⇒ tan α = 4 3 =⇒ arctan(4 3 ) = α.
For θ = arg(z), we can say that
=⇒ θ = α 3 = 1 3 arctan(4 3 ).
There are 3 roots:
=⇒ z k = 3√ 5(cos(θ + 2kπ
2kπ
) + i sin(θ +
3 3 ))
k = 0 =⇒ z 0 = 3√ 5(cos θ + i sin θ)
k = 1 =⇒ z 1 = 3√ 5(cos(θ + 2π 3 ) + i sin(θ + 2π 3 ))
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