If you can solve these complex number questions correctly, then show it!?

Question

1) Write the complex number (sqrt(3)/2)-(1/2i) in the form re^i(theta).

2) Write the complex number 2e^(-pi/4)i in the form a + bi.

3) Write the complex number e^(3+4i)t in the form a(t)+b(t)i.

Answer 1

1) (√3)/2 - (1/2) i
= cos π/6 - i sin π/6
= cos (-π/6) + i sin (-π/6)
= e^(-iπ/6).

2) 2e^(-iπ/4)
= 2 (cos (-π/4) + i sin (-π/4))
= 2 (1/√2 + i (-1/√2))
= √2 (1 - i)
[= √2 - √2 i].

3) e^((3+4i)t)
= e^(3t + i.4t)
= e^(3t) . e^(4ti)
= e^(3t) . (cos 4t + i sin 4t)
[= e^(3t) cos 4t + (e^(3t) sin 4t) i].

In the last two cases the final line is enclosed in brackets to indicate that it's there only to conform exactly with the requested format - I'd normally leave it in the form of the previous line.

Answer 2

The formulas to transform a complex number in the form a + bi into the form re^i(theta) are:
r = sqrt(a^2 + b^2)
theta = tan^-1(b / a)

The formulas to transform a complex number in the form re^i(theta) into the form a + bi are:
a = r * cos theta
b = r * sin theta


1) r = sqrt[(sqrt(3) / 2)^2 + (-1/2)^2] = sqrt(3/4 + 1/4) = 1
theta = tan^-1 [(-1/2) / (sqrt(3)/2] = 2 pi / 3
(for theta, either use a calculator or look it up in a trig table)
The answer is 1e^i(2pi / 3)

2) a = 2 * cos(-pi / 4) = sqrt(2)
b = 2 * sin(-pi / 4) = -sqrt(2)
The answer is sqrt(2) - sqrt(2) i

3) First use exponent law to change the expression into
(e^3)e^i(4t)
Now a = (e^3) cos(4t)
b = (e^3) sin(4t)
The answer is (e^3) cos(4t) + (e^3) sin(4t) i

Answer 3

one million.25 = 4.25 * sqrt(3 hundred/L) sqrt(3 hundred/L) = one million.25 / 4.25 sqrt(3 hundred/L) = 5/17 sq. the two aspects: 3 hundred/L = (5/17)^2 3 hundred/L = 25/289 25L = 3 hundred*289 25L = 86700 L = 86700 / 25 L = 3468