Finding z^4 in Polar & Cartesian Forms

[menco]

Homework Statement

Express z=-1+4i in polar for then find z^4 converting to Cartesian form

Homework Equations

r = sqrt(x^2+y^2)
theta = y/x
z= r cos (theta) + i r sin (theta)

The Attempt at a Solution

r= sqrt(-1^2+4^2)
= sqrt(17)

theta = tan a = 4/1
a = tan^-1 (4/1)
= 1.3258
pi-a = 1.8158 (to find the argument from the real axis)

in polar form = sqrt(17) cis 1.8158

Then to cartesian form

z^4 = r^4 cos 4(theta) + i r^4 sin 4(theta)
= (sqrt (17))^4 cos 4(1.8158) + i (sqrt(17))^4 sin 4(1.8158)
= 289 cos 7.2632 + i 289 sin 7.2632
z = 4sqrt(160.9760) + 4sqrt(240.0161) i
z = 3.5620 + 3.9360 i

Im not sure if this is correct but hopefully i am on the right track. I was not sure if I was converting to Cartesian form correctly or if i should introduce z^4 at the polar form so that z^4 = r^4 cis 4(theta)

 

[Mentallic]

z^4 = 289 cos 7.2632 + i 289 sin 7.2632
z = 4sqrt(160.9760) + 4sqrt(240.0161) i
z = 3.5620 + 3.9360 i

This is where you went wrong. You had found z⁴ correctly, but then in order to convert back to z
(which you shouldn't be doing anyway because the question asked you to convert z⁴ back into Cartesian form, not z)
you should have simply calculated 289\cdot \cos(7.2632) and 289\cdot \sin(7.2632)

Now something else you should work on is avoiding the decimal approximations. If you calculate what you have, you won't get the exact answer. For example, 289\cdot \cos(7.2632)+i\cdot289\cdot\sin(7.2632)\approx 160.976+240.016i but the exact answer to the problem is \left(-1+4i\right)^4=161+240i

So let \alpha = \tan^{-1}(4) therefore the argument is \pi-\alpha and so we have

z=-1+4i=\sqrt{17}\left(\cos(\pi-\alpha)+i\sin(\pi-\alpha)\right)

And we can simplify these cos and sin expressions, and after that, we can then find z⁴. Now, can you simplify the expressions \cos(\tan^{-1}\theta) and \sin(\tan^{-1}\theta) ? And also, what about expressing \cos(4x) and \sin(4x) in terms of \sin(x) and \cos(x)?

 

[grzz]

a = tan^-1 (4/1)
= 1.3258
pi-a = 1.8158 (to find the argument from the real axis)

I think that the argument required for z (which according to convention is measured from the real axis in an anticlockwise manner) is just a = tan^-1(4/1) and not (∏ - a).

 

[grzz]

I think that the argument required for z (which according to convention is measured from the real axis in an anticlockwise manner) is just a = tan^-1(4/1) and not (∏ - a).

I am so sorry!

I did not notice the -1 in z!

 

[Mentallic]

grzz, you can go back and edit your first post if you made a mistake somewhere. Better than making a second post