# Complex Numbers [Lecture notes] by Peter M. Neumann

By Peter M. Neumann

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Extra info for Complex Numbers [Lecture notes]

Example text

1 Algebraic Representation of Complex Numbers (b) (c) (d) (e) (f) 21 |z| + z = 3 + 4i; z 3 = 2 + 11i, where z = x + yi and x, y ∈ Z; iz 2 + (1 + 2i)z + 1 = 0; z 4 + 6(1 + i)z 2 + 5 + 6i = 0; (1 + i)z 2 + 2 + 11i = 0. 23. Find all real numbers m for which the equation z 3 + (3 + i)z 2 − 3z − (m + i) = 0 has at least one real root. 24. Find all complex numbers z such that z = (z − 2)(z + i) is a real number. 1 25. Find all complex numbers z such that |z| = | |. z √ 26. Let z1 , z2 ∈ C be complex numbers such that |z1 + z2 | = 3 and |z1 | = |z2 | = 1.

Note that if λ > 0, then the vectors λ− v and − v have the same orientation and → → |λ− v | = λ|− v |. → − When λ < 0, the vector λ v changes to the opposite orientation, and → − → → → |λ− v | = −λ|− v |. Of course, if λ = 0, then λ− v = 0 (Fig. 9). 9. Examples. (1) We have 3(1 + 2i) = 3 + 6i; therefore, M (3, 6) is the geometric image of the product of 3 and z = 1 + 2i. (2) Observe that −2(−3 + 2i) = 6 − 4i; we obtain the point M (6, −4) as the geometric image of the product of −2 and z = −3 + 2i (Fig.

2). 2, the geometric image √ P1 (−1, −1) lies in the third quadrant. Then r1 = (−1)2 + (−1)2 = 2 and t∗1 = arctan π 5π y + π = arctan 1 + π = + π = . 2. Hence z1 = √ 5π 5π + i sin 2 cos 4 4 and Arg z1 = 5π + 2kπ|k ∈ Z . 4 (b) The point P2 (2, 2) lies in the ﬁrst quadrant, so we can write r2 = Hence √ π 22 + 22 = 2 2 and t∗2 = arctan 1 = . 4 √ π π z2 = 2 2 cos + i sin 4 4 and Arg z = (c) The point P3 (−1, π + 2kπ|k ∈ Z . 4 √ 3) lies in the second quadrant, so (Fig. 3) √ 2π π . 3. and Arg z3 = 2π + 2kπ|k ∈ Z .