# There are 7 questions in the attached file about polar coordinates and roots need to be solved

*label*Science

*timer*Asked: Feb 4th, 2019

*account_balance_wallet*$9.99

### Question Description

please type the answers in steps.do your best please.

## Tutor Answer

Here is the assignment :).. Allow me to me know if you have questions or need another help okay ;)

QUESTIONS:

Answer for Number 1:

Solution:

For ππ = π + ππ:

The polar of a complex number is: 2 = π ( πππ π + π π πππ)

π

Where π = βπ2 + π 2 and π = π‘ππβ1 (π) π > 0

π

= π‘ππβ1 (π) + 1000 π < 0, π > 00π < 0

π

=π‘ππβ1 (π) + 3600 , π > 0, π > 0

r = β(1)2 + (2)2 = β5

2

π = π‘ππβ1 (1) = π‘ππβ1 2 = 1.107

r = 63.3%

Hence, the polar for of π§1 = 1 + 2π would be β5(cos(1.107) + π sin(1.107)

The polar coordinates would be: βπ ππ¨π¬(π. πππ) , βπ π¬π’π§(π. πππ)

For ππ = βπ + ππ

r = β(β3)2 + (4)2 = β5

4

ΞΈ = π‘ππβ1 (β3) + 180

= β53.13 + 180

= 126.860 = 2.21 ππππππ

= 5(cos(126.860 ) + π sin(126. 860 ))

Hence, the polar coordinates would be: π(ππ¨π¬( πππ. πππ ) + π π¬π’π§(πππ. πππ ))

For ππ = βπ β ππ

r = r = β(5)2 + (6)2 β β25 + 36 = 7.81

β6

ΞΈ = π‘ππβ1 ( 5 ) + 180

= 50.19 + 180

= 230....

*flag*Report DMCA

Brown University

1271 Tutors

California Institute of Technology

2131 Tutors

Carnegie Mellon University

982 Tutors

Columbia University

1256 Tutors

Dartmouth University

2113 Tutors

Emory University

2279 Tutors

Harvard University

599 Tutors

Massachusetts Institute of Technology

2319 Tutors

New York University

1645 Tutors

Notre Dam University

1911 Tutors

Oklahoma University

2122 Tutors

Pennsylvania State University

932 Tutors

Princeton University

1211 Tutors

Stanford University

983 Tutors

University of California

1282 Tutors

Oxford University

123 Tutors

Yale University

2325 Tutors