Products, quotients and powers in trigonometric form, and the exponential form
Answers are for parents and teachers.
1
(a) Divide 6(cos 7π/10 + i sin 7π/10) by 3(cos π/5 + i sin π/5). (b) Multiply 2(cos 5π/6 + i sin 5π/6) by 3(cos 5π/6 + i sin 5π/6) and give the answer with principal amplitude, as x + yi.
(a) 2(cos(7π/10 − 2π/10) + i sin(7π/10 − 2π/10)) = 2(cos π/2 + i sin π/2) = 2i. (b) Modulus 6, amplitude 5π/3, which is −π/3. So 6(cos(−π/3) + i sin(−π/3)) = 3 − 3√3 i.
2
Find (√3 − i)⁵ as x + yi.
√3 − i = 2(cos(−π/6) + i sin(−π/6)). The fifth power has modulus 32 and amplitude −5π/6: 32(−√3/2 − i/2) = −16√3 − 16i.
3
Write in exponential form (a) −3 + 3√3 i (b) 4 − 4i (c) −7i.
(a) r = 6, reference angle π/3, quadrant II: 6e^(2πi/3). (b) r = 4√2, amplitude −π/4: 4√2 e^(−iπ/4). (c) 7e^(−iπ/2).
4
Let Z1 = 1 + √3 i and Z2 = √2 − √2 i. Find Z1Z2 and Z1 ÷ Z2 in exponential form.
Z1 = 2e^(iπ/3) and Z2 = 2e^(−iπ/4). Z1Z2 = 4e^(i(π/3 − π/4)) = 4e^(iπ/12). Z1 ÷ Z2 = e^(i(π/3 + π/4)) = e^(7πi/12).
5
Find ((1 + √3 i)/(1 − √3 i))³.
1 + √3 i = 2e^(iπ/3) and 1 − √3 i = 2e^(−iπ/3), so the quotient is e^(2πi/3). Cubed: e^(2πi) = 1.