MCS 351 ENGINEERING MATHEMATICS SOLUTION OF ...

CHAPTER 13 - COMPLEX NUMBERS AND FUNCTIONS
13.1
1. We can write −i as 0 + i(−1). Then,
• i 2 = [0 + i(−1)][0 + i(−1)] = 0 − 1 + i.(0 + 0) = −1,
• i 3 = i 2 .i = (−1).i = −i,
• i 4 = i 3 .i = (−i).i = −i 2 = −(−1) = 1,
• i 5 = i 4 .i = 1.i = i
We know that i = −i.
• 1 i = −i
−i 2
= −i
1 = −i,
• 1 i 2 = 1
−1 = −1,
• 1 i 3 = 1 −i = i
−i 2 = i 1 = i.
2. Let z = 2 + 2i. Then we obtain iz = i(2 + 2i) = −2 + 2i.
Let z = −1 − 5i. Then iz = i(−1 − 5i) = 5 − i.
Let z = 4 − 3i. Then iz = i(4 − 3i) = 3 + 4i.
a + b = 90 ◦
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