MCS 351 ENGINEERING MATHEMATICS SOLUTION OF ...

2. z(t) = 5 − 2it, −3 ≤ t ≤ 3 ⇒ x(t) = 5, y(t) = −2t
• t = −3 ⇒ x(−3) = 5, y(−3) = 6 ⇒ starting point:(5, 6)
• t = 3 ⇒ x(3) = 5, y(3) = −6 ⇒ ending point:(5, −6)
Then we obtain figure below:
3. z(t) = z 0 + re it , 0 ≤ t ≤ 2π denotes the circle of radius r with center z 0 .
For z(t) = 4 + i + 3e it , we find z 0 = 4 + i and r = 3.
z(t) = 4 + i + 3(cos t + i sin t) = (4 + 3 cos t) + i(1 + 3 sin t) ⇒ x(t) = 4 + 3 cos t, y(t) = 1 + 3 sin t
For finding orientation, we should determine z(t) at some points.
• t = 0 ⇒ z(0) = 4 + i + 3e 0 = 7 + i
• t = π ⇒ z(π) = 4 + i + 3e iπ = 4 + i + 3(cos π + i sin π) = 4 + i + 3(−1 + i.0) = 1 + i
• t = 2π ⇒ z(2π) = 4 + i + 3e i2π = 4 + i + 3(cos 2π + i sin 2π) = 4 + i + 3(1 + i.0) = 7 + i
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