Question:

The principal value of the arg(z)arg (z) and z | z | of the complex number z=1+cos(11π9)+isin11π9z=1+\cos\left(\frac{11\pi}{9}\right)+ i \, \sin\frac{11\pi}{9} are respectively

Updated On: Jul 2, 2022
  • 11π8,2cos(π18)\frac{11\pi}{8},2\,\cos\left(\frac{\pi}{18}\right)
  • 7π18,2cos(11π18)-\frac{7\pi}{18},-2\,\cos\left(\frac{11\pi}{18}\right)
  • 2π9,2cos(7π18)\frac{2\pi}{9},2\,\cos\left(\frac{7\pi}{18}\right)
  • π9,2cos(π18)-\frac{\pi}{9},-2\,\cos\left(\frac{\pi}{18}\right)
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

z=2cos211π18+2isin11π18cos11π18z=2\,cos^{2} \frac{11\pi}{18}+2i\,sin \frac{11\pi}{18}\, cos\, \frac{11\pi}{18} =2cos11π18cis(11π18)=2\,cos\, \frac{11\pi}{18}\, cis \left(\frac{11\pi}{18}\right) But 11π18\frac{11\pi}{18} is in the IlndIlnd quadrant cos11π18<0\therefore cos \frac{11\pi}{18} < 0 z=2cos(11π18)cis(11π18π)\therefore z=-2\,cos\left(\frac{11\pi}{18}\right)cis\left(\frac{11\pi}{18}-\pi\right) =2cos(11π18)cis(7π18)= -2\,cos\left(\frac{11\pi}{18}\right)cis\left(-\frac{7\pi}{18}\right) Argz=7π18\therefore Arg z=-\frac{7\pi}{18} i.e., z=2cos(π18)\left|z\right|=-2\,cos\left(\frac{\pi}{18}\right)
Was this answer helpful?
0
0

Concepts Used:

Complex Numbers and Quadratic Equations

Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.

Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.