express -6+6i into polar form?

Answer 1

The magnitude is sqrt(6^2+6^2)=sqrt(72)
the angle is arctan(-1)=225 degrees.

Answer 2

This means that , from the origin, go 6 to the left and then 6 vertically and join points to obtain a right angled triangle.
Hypotenuse = H, say
Thus H² = 6² + 6² = 72
H = 8.49
To obtain angle that H makes with the -X axis, use tanß = 6/6 = 1 so ß =45°
Angle made with X axis is then 225°
In polar form -6 + 6i = 8.49 angle 225°
which can be written as 8.49(cos225° + isin225°)

Answer 3

For those type of equations or kinds, continually draw a precise triangle and call one side (often contained in the cartesian plane, the x axis) -6?3 and the different side 6. So the hypoteneous is the cost and the attitude is the inverse tangent of imaginary section divided with information from the real section. for this reason, the cost is: d = ( ( -6?3 )^2 + ( 6 )^2 ) ^0.5 = 12 The angles is: theta = (y/x) = arctan ( 6 / -6?3) = -30 levels for this reason, the polar kind is: 12<-30 levels. So, contained in the Cartesian plane (i.e. in graphing diagrams) the y = 6 and the x = -6?3. desire this helps.

Answer 4

To find r, given the value of z=a+bi, compute |z| = sqrt (a^2 + b^2). To compute theta, remember that tan theta = y/x, and because z in the second quadrant (a<0 & b>0), theta should be between pi/2 and pi.* Finally, z = r cis theta.

~~~
*But if z was in the first quadrant (a, b > 0), theta would be between 0 and pi/2, etc. For the special case that z is on the positive y-axis (b>0 & a=0), theta = pi/2, and for the special case that z is on the negative axis, theta = 3*pi/2, and for the third special case of the origin, just let theta = 0 arbitrarily, even though it can be anything, here.

Answer 5

-6+6i is in the second quadrant.
r = 6√2
θ = arctan (-1) = 135 degree
-6+6i = 6√2e^(135 i+360n i)

Answer 6

I got 6 root 2 cis 3Pi/4

Answer 7

r = SQRT[(-6)^2 + 6^2] = SQRT 72
tan x = 6/(-6)
=> x = tan-1(-1)
=> x = -45, -45+180
=> x = 135 (for 0
r = (SQRT 72) cos135 + i[(SQRT 72)sin 135]