Given w=2+2i and v=−5√3+5i,
vw^2 can be expressed as r*(cos(α∘)+i*sin(α∘)),
where r is a real number and 0≤α≤360. What is the value of r+α?
i found
-40i(sqrt(3)-i)
-80i(1/2sqrt(3) -i)
80(1-i*sqrt(3)/2)
then i found that r= 80 and α is -120∘
but r+a must be positive integer
where is the wrong one
vw^2 = (2√2,45°)(2,150°)^2
= (2√2,45°)(4,300°)
= (8√2,345°)
To find the value of vw^2, we need to first calculate the product of w and w, and then multiply the result by v.
1. Find the product of w and w:
w * w = (2 + 2i) * (2 + 2i)
Using the FOIL method (First, Outer, Inner, Last):
w * w = 2 * 2 + 2 * 2i + 2i * 2 + 2i * 2i
= 4 + 4i + 4i - 4
= 4 + 8i - 4
= 8i
2. Multiply the result by v:
vw^2 = (8i) * (-5√3 + 5i)
Apply the distributive property:
vw^2 = -40√3i - 40i^2 + 40i^2
Simplify:
vw^2 = -40√3i
So, vw^2 can be expressed as -40√3i.
To express it in the desired form, r*(cos(α∘) + i*sin(α∘)), we can set r = -40√3 and α = 90°.
Therefore, r + α = -40√3 + 90 = -2√3 + 90.