Write each expression in the standard form for a complex number, a + bi.
A. [3(cos(27°)) + isin(27°)]^5
B. [2(cos(40°)) + isin(40°)]^6
For A i got 2.67+1.36i and for B i got 1.53+1.29i
(3cis27°)^5
= 3^5 cis135°
= 243 (1/√2 - 1/√2 i)
(2cis40°)^6
= 2^6 cis240°
= 64 (-1/2 - √3/2 i)
you can make those into a+bi form, but you need to review deMoivre's formulas
To write each expression in the standard form for a complex number (a + bi), we need to simplify the expression using De Moivre's theorem.
Let's start with expression A: [3(cos(27°)) + isin(27°)]^5.
1. Start by using De Moivre's theorem: (r(cosθ + isinθ))^n = r^n(cos(nθ) + isin(nθ)).
In this case, r = 3, θ = 27°, and n = 5.
2. First, calculate r^n: 3^5 = 243.
3. Now, calculate nθ: 5 * 27° = 135°.
4. Convert 135° to radians: 135° * π/180° = 3π/4.
5. Plug in the values in the formula: [3(cos(27°)) + isin(27°)]^5 = 243(cos(3π/4) + isin(3π/4)).
6. Simplify the expression inside the parentheses:
a. cos(3π/4) = cos(45°) = 1/√2 ≈ 0.707
b. sin(3π/4) = sin(45°) = 1/√2 ≈ 0.707
Therefore, [3(cos(27°)) + isin(27°)]^5 = 243(0.707 + 0.707i).
So, the complex number in standard form for expression A is approximately 170.301 + 170.301i.
Now, let's move on to expression B: [2(cos(40°)) + isin(40°)]^6.
1. Using De Moivre's theorem, we have r = 2, θ = 40°, and n = 6.
2. Calculate r^n: 2^6 = 64.
3. Calculate nθ: 6 * 40° = 240°.
4. Convert 240° to radians: 240° * π/180° = 4π/3.
5. Apply the formula: [2(cos(40°)) + isin(40°)]^6 = 64(cos(4π/3) + isin(4π/3)).
6. Simplify the expression inside the parentheses:
a. cos(4π/3) = cos(240°) = -1/2
b. sin(4π/3) = sin(240°) = -√3/2
Therefore, [2(cos(40°)) + isin(40°)]^6 = 64(-1/2 - (√3/2)i).
The complex number in standard form for expression B is -32 - 32√3i.
Keep in mind that these are approximations, and you may want to round the values depending on the required level of precision.