SOLUTION: Choose the correct quadrant to use in this conversion from standard to trigonometric form: 1) 6-11i I am having a very difficult time understanding this question. Any help is

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Question 394201: Choose the correct quadrant to use in this conversion from standard to trigonometric form:
1) 6-11i
I am having a very difficult time understanding this question. Any help is greatly appreciated.
Found 2 solutions by josmiceli, Edwin McCravy:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
(6- 11i) is a point in the complex plane. The real part, 6
is plotted on the horizontal axis, and the imaginary part, -11i,
is plotted on the vertical axis.
If you were to draw the axes, measure 6 units to the right and
11 units down (since it's negative), then you have plotted a point
in the 4th quadrant.
----------------------
In order to express this in trig form, you need an amplitude
and an angle
The angle is a positive angle that is 270 degrees (3 quadrants) +
the angle whose tangent is 6/11.
6%2F11+=+.5454
tan^-1(.5454) = 28.61 degrees
270 + 28.61 = 298.61 degrees
The amplitude is sqrt%286%5E2+%2B+11%5E2%29
sqrt%2836+%2B+121%29+=+sqrt%28157%29
sqrt%28157%29+=+12.53
So this is a line 12.53 units from the origin at an angle of 298.61 degrees
in the + direction (counterclockwise)

Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!

6-11i

Rule:

A + Bi is represented by the line segment that goes from the origin to
the point (A,B).  It has length r.

So,
6 - 11i is represented by the line segment that goes from the origin to
the point (6,-11)

So we draw that line segment and label it r in length:



Next we'll indicate the angle q starting
at the right hand of the x-axis going around counter-clockwise
to the line we just drew.  I'll indicate q with
a red arc: 



Next we draw a line from that point perpendicular to the x-axis.
I'll draw it in green:



That makes a right triangle, so we label the horizontal leg
as the x value of the point. That is, x = 6.  We label the
vertical leg the y -value of the point. That is y = -11 



Next we calculate the value of r using the Pythagorean theorem:

r%5E2=x%5E2%2By%5E2
r%5E2=%286%29%5E2%2B%28-11%29%5E2
r%5E2=36%2B121
r%5E2=157
r=sqrt%28157%29

So we label the r as r = sqrt%28157%29



Next we calculate q

by first calculating the tangent of the reference angle,

and then placing it in the 4th quadrant.

tan%28theta%29=y%2Fx
tan%28theta%29=-11%2F6
Use the inverse tangent function on the calculator to find
the inverse tangent of %2B11%2F6 to get the reference angle:

reference angle = 61.38954033°

To get q in the 4th quadrant subtract
the reference angle from 360° and get 

q = 298.6° rounded to the nearest
tenth of a degree.

So we label q = 298.6°



Now the trigonometric form is

r(cosq + i·sinq)

and upon substituting, the final answer is

sqrt%28157%29(cos298.6° + i·sin298.6°)

Edwin