Download Complex Numbers: Plotting and Representing in the Complex Plane - Prof. Jonathan Comes and more Assignments Mathematics in PDF only on Docsity! Math 112 Jonny Comes Fall 2005 Assignment #9 Partial Solutions Additional Exercises: (Be sure to justify all your answers) 1. Plot each of the following complex numbers in the complex plane, and write each of the complex numbers in standard form (a + bi). (a) eπi Solution: Here’s the picture: 1 π Now to put eπi into standard form: eπi = r cos θ + ir sin θ = 1 cos π + i · 1 sinπ = −1 (b) e π 2 i Solution: Here’s the picture: 1 π 2 Now to put e π 2 i into standard form: e π 2 i = r cos θ + ir sin θ = 1 cos π 2 + i · 1 sin π 2 = i (c) 3e π 4 i Solution: Here’s the picture: 3 π 4 Now to put 3e π 4 i into standard form: 3e π 4 i = r cos θ + ir sin θ = 3 cos π 4 + i · 3 sin π 4 = 3 (√ 2 2 ) + 3i (√ 2 2 ) = 3 √ 2 2 + 3 √ 2 2 i (d) 6e 2π 3 i Solution: Here’s the picture: 6 2π 3 Now to put 6e 2π 3 i into standard form: 6e 2π 3 i = r cos θ + ir sin θ = 6 cos 2π 3 + i · 6 sin 2π 3 = 6 ( −1 2 ) + 6i (√ 3 2 ) = −3 + 3 √ 3i Now to put 5 + 5i into polar form we first notice r = √ a2 + b2 =√ 52 + 52 = √ 50 = 5 √ 2. Now since arctan(b/a) = arctan(5/5) = arctan(1) = π4 and this agrees with the picture we see θ = π 4 [or you can just see from the picture that θ = π4 ]. So 5 + 5i = 5 √ 2e π 4 i (d) 2 − 2 √ 3i Solution: Here’s the picture: -2 2 3 Now to put 2−2 √ 3i into polar form we first notice r = √ a2 + b2 = √ 22 + (−2 √ 3)2 = √ 4 + 12 = √ 16 = 4. Now since arctan(b/a) = arctan(−2 √ 3/2) = arctan(− √ 3) = −π3 and this agrees with the picture we see θ = −π3 . So 2 − 2 √ 3i = 4e− π 3 i (e) − √ 3 + i Solution: Here’s the picture: 1 3 Now to put − √ 3+i into polar form we first notice r = √ a2 + b2 = √ (− √ 3)2 + 12 = √ 3 + 1 = √ 4 = 2. Now arctan(b/a) = arctan(−1/ √ 3) = −π6 , but this does not agree with the picture so θ = arctan(b/a)+ π = −π6 + π = 5π6 . So − √ 3 + i = 2e 5π 6 i 3. Write the following in Euler e-notation (reiθ). (a) ( 2eπi ) ( 5e− π 2 i ) Solution: ( 2eπi ) ( 5e− π 2 i ) = 2 · 5eπi−π2 i = 10eπ2 i (b) (√ 2e π 3 i )(√ 2e π 6 i ) Solution: (√ 2e π 3 i )(√ 2e π 6 i ) = √ 2 · √ 2e π 3 i+π 6 i = 2e 3π 6 i = 2e π 2 i (c) 6e 3π 2 i 2e− π 2 i Solution: 6e 3π 2 i 2e− π 2 i = 6 3 e 3π 2 i−(−π 2 i) = 2e 4π 2 i = 2e2πi = 2 (d) 7e π 6 i 21e 2π 3 i Solution: 7e π 6 i 21e 2π 3 i = 7 21 e π 6 i− 2π 3 i = 1 3 e− 3π 6 i = 1 3 e− π 2 i 4. Plot the points −iz, −iw and −iu where z, w, and u are the complex numbers pictured below u w Im Re z Solution: Because −i = e−π2 i, multiplication by −i is just a rotation by −π/2 radians (or 90◦ clockwise) so here is the picture: u w Im Re z - i z - i u - i w 5. Plot the two points z̄ and z · z̄ where z is the complex number pictured below. z Im Re 2
