Maths help. How can I change below expression into polar form and standard form using Euler's formula?

Question

((1+ i)/(root3 + i)) all to power 6

Answer 1

1+i = an angle of 45 degrees or pi/4 radians, so 1+i = e^[(pi/4)*i]. {e^[(pi/4)*i]}^6 = e^[(3pi/2)*i]. 3pi/2 radians = 270 degrees --> = 0-i.

Now do the same for the bottom. sqrt(3)+i --> e^[(pi/6)*i]. Continue from here.

Answer 2

The approach used by the first answerer is okay (that is, convert the numerator and the denominator to polar form separately), but he has an error. It's true that the polar form of 1+i has angle pi/4, but the magnitude of e^(i*pi/4) is 1. (In fact, the magnitude of e^i*x is 1 for any real x.)

The magnitude of (1+i) is sqrt(1^2 + 1^2) = sqrt(2). So

1+i = sqrt(2) e^(i*pi/4)

The same correction needs to be made for the denominator calculation.

Answer 3

Jeez..tuff one mate..was off the day they did functions of complex variables and representations of mixed numbers on the argand plane etc..but will have a go..
mult top and bottom by complex conjugate..
>(1+i)(rt3-i)/4> (rt3+1)/4+(i(rt3-1)/4)
this is now in form a+ib where a=(r3+1)/4 and b=(rt3-1)/4
in polar form this corr to z=rcos(alpha)+irsin(alpha)
and Euler says this is just z=re^(ialpha) where alpha =tan-1(b/a)..
so z^6= r^6e^(6ialpha)
[z]=rt(a^2+b^2)=0.5 and alpha=cos-1((rt3-1)/4)/.5))= 0.38pi
so z^6=(.5)^6(cos(2.28pi)+ isin(2.28pi)
so z^6=(.5)^6(x+iy) where x=cos 2.28pi and y= sin2.28pi
Hope this helps.. did it in my head from memory!

Answer 4

The equation of a line in slope intercept sort is y=mx+b and the equation for widely used sort is Ax+via employing=C. So if the customary sort equation is y=3x+9 to get it into satndard sort you subtract 3x from the two aspects. a thank you to do the transformation is to get the equation so as that x and y alongside with their coefficients are on one section and there is a single sort at different. be effective to checklist the x first then the y. in case you go with for help with a specufic subject, enable me understand