Stitz-Zeager_College_Algebra_e-book

Instead we opt for a more direct approach we

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Unformatted text preview: (z ) = 0 2 2 2 and Im(z ) = 1. Since i is called the ‘imaginary unit,’ these answers make perfect sense. 2. To write a polar form of a complex number z , we need two pieces of information: the modulus |z | and an argument (not necessarily the principal argument) of z . We shamelessly mine our solution to Example 11.7.1 to find what we need. √ (a) For z = 3 − i, |z | = 2 and θ = − π , so z = 2cis − π . We can check our answer by 6 6 √ converting it back to rectangular form to see that it simplifies to z = 3 − i. √ √ (b) For z = −2 + 4i, |z | = 2 5 and θ = π − arctan(2). Hence, z = 2 5cis(π − arctan(2)). It is a good exercise to actually show that this polar form reduces to z = −2 + 4i. (c) For z = 3i, |z | = 3 and θ = π . In this case, z = 3cis π . This can be checked 2 2 geometrically. Head out 3 units from 0 along the positive real axis. Rotating π radians 2 counter-clockwise lands you exactly 3 units above 0 on the imaginary axis at z = 3i. (d) Last but not least, for z = −117, |z | = 117 and θ = π . We get z = 117cis(π ). As with the previous problem, our answer is easily checked geometrically. 848 Applications of Trigonometry T...
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