QUESTION: simplify a complex number and give the polar form. (1+i)^5 (square root of 3-i)^-4
(1+I)^5 = 1 + 5i + 10i^2 + 10i^3 + 5i^4 + i^5
(1+I)^5 = 1 + 5i - 10 - 10i + 5 + i = -4 - 4i
(square root of 3-i)^-4 = (sqrt(3-i))^(-4) = ((3-i)^(1/2))^(-4) = (3-i)^(-2)
Now first of all,
(3-i)^2 = 9 - 6i - 1 = 8 - 6i
So,
(3-i)^(-2) = (1/(8 - 6i)) = (8 + 6i)/[(8 - 6i)(8 + 6i)] = (8 + 6i)/(64 + 36) = (8 + 6i)/100 = (4 + 3i)/50
Finally,
(1+I)^5 * (sqrt(3-i))^(-4) = (-4 - 4i)*(4 + 3i)/50 = (-16 - 12i - 16i + 12)/50 = (-4 - 28i)/50 = (-2 - 14i)/25
The complex number is z = -2/25 - 14i/25
In polar form, z = r.e^(it) = r(cos(t) + i.sin(t))
So, r.cos(t) + i.r.sin(t) = -2/25 - 14i/25
Then
r.cos(t) = -2/25
r.sin(t) = -14/25
And,
tan(t) = sin(t)/cos(t) = (-14/25)/(-2/25) = 7
Therefore, t = atan(7) = 1.4289 rad
And, (r.cos(t))^2 + (r.sin(t))^2 = r^2(cos^2(t) + sin^2(t)) = r^2
So, (-2/25)^2 + (-14/25)^2 = r^2
r^2 = (4 + 196)/(25)^2 = 200/(25)^2 = 8/25
So, r = rt(8)/5 = 2rt(2)/5
z = r(cos(t) + i.sin(t)), where r = 2rt(2)/5 and t = 1.4289 rad