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Unformatted text preview: more, we have tan θ = 1. Next, draw the complex √ number z = −1 − i on the Argand diagram. Clearly the equations (7) are satisﬁed with r = 2 and θ = −3π/4. Furthermore, we again have tan θ = 1. Note that the equation θ = tan−1 1 has two solutions for θ in the range −π < θ ≤ π . (2) Suppose that y = 0, so that z = x ∈ R. Then |x| = √ x2 = x if x ≥ 0, −x if x < 0, and this is simply the absolute value of the real number x. (3) In view of Remark (1) above, we need to exercise extreme care when we try to determine an angle θ which satisﬁes the equations (7). The best advice is always to place the complex number z on the Argand diagram and determine ﬁrst of all a suitable range for θ. For example, we know that if z = −1 − i,
Chapter 1 : The Number System page 9 of 19 First Year Calculus c W W L Chen, 1982, 2005 then a suitable range for θ may be π < θ < 3π/2 or −π < θ < −π/2. Once such a suitable range is determined, the equa...
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