Heres what I got But I really would like to see the correctway to break this all
ID: 3094899 • Letter: H
Question
Heres what I got But I really would like to see the correctway to break this all down or even finish the problem..Here itis.... z = 1 + i sqrt (3)CONVERT TO COMPLEX #
or
a a + bj
FIRST..
sqrt 3 = 1.732
REWRITE PROBLEM
1 + i (1.732)
CONVERT TO POLAR WHERE..
r = sqrt 1 + sqrt 1.732
OR..
r = 1^2 + 1.732^2
ANSWER IS..
3.999824 ROUND TO 4
RADIUS IS THE SQRT OF 4
sqrt4 = 2
And i managed to confuse myself cause wheres the i ??
This is far as i got cause every time I try to use arc sin tofigure
1.732/2 I get answer of "overflow"
And havent quite figured out the other way yet.....
Heres what I got But I really would like to see the correctway to break this all down or even finish the problem..Here itis.... z = 1 + i sqrt (3)
CONVERT TO COMPLEX #
or
a a + bj
FIRST..
sqrt 3 = 1.732
REWRITE PROBLEM
1 + i (1.732)
CONVERT TO POLAR WHERE..
r = sqrt 1 + sqrt 1.732
OR..
r = 1^2 + 1.732^2
ANSWER IS..
3.999824 ROUND TO 4
RADIUS IS THE SQRT OF 4
sqrt4 = 2
And i managed to confuse myself cause wheres the i ??
This is far as i got cause every time I try to use arc sin tofigure
1.732/2 I get answer of "overflow"
And havent quite figured out the other way yet.....
Explanation / Answer
Let z = a + bi. Then r = sqrt(a^2+b^2) = sqrt(1^2+(sqrt3)^2) = 2 and phi = arctan(b/a) = arctan(sqrt3) = 60° = pi/3 So you have polar form: z = r*e^(i*phi) = 2*e^(i*pi/3) and trigonometric form: z = r*(cos(phi) + i*sin(phi)) = 2*(cos(pi/3) + i*sin(pi/3))