# Complex numbers help

## Homework Statement

1.
(a) Express -1 + √3i in modulus-argument form. Evaluate (-1 + √3i)^8
expressing your answer in (a + ib) form.

Find also the square roots of -1 + √3i in (a + ib) form.
(b) Use complex numbers to find

(intergral is between 0 and ∞) ∫ e^-x cos2x dx.

2.
(a) Find the modulus and argument of -1 - i√3. Hence find (-1 - i√3) ^ 10 in a+ib
form. Find also the square roots of -1 - i√3 in a + ib form.

(b) Use complex numbers to find

∫ e^-x sin3x dx

3.
(a) Find the square roots of 1 - i√3. Find also
[(1- i√3)/ (1+i√3)] ^ 8

(b) Use complex numbers to find

∫ e^kx cosx dx

where k is a constant.

## The Attempt at a Solution

For q1. but nt sure u gta check it for me q2 seems similar but q3 lost on it.

q1.
1. Modulus = √ ((-1)&2 + (√3)^2)
= √(1 + 3) = 2
Argument = arctan(√3/-1) = 4 pi /3 (draw a picture to make sure you have the angle in the correct quadrant)

The 8th power has modulus 2^8 = 256 and argument 8 x (4 pi/3) = 32 pi /3 = 2pi/3 (subtract multiples of 2 pi).

modulus 2 pi /3 = cos (2pi/3) + i sin (2pi/3) = -0.5 + √3/2 i

(-1 + √3i)^8 = 256 (-0.5 + √3/2 i) = -128 + 128√3 i

The square root is similar - modulus √2, amplitude 2 pi / 3

(b) ∫ e^-x sin3x dx = Im ∫ e^-x(cos 3x + i sin 3x) dx
= Im ∫ e^x e^3ix dx
= Im ∫ e^(-1+3i) x dx
= Im e^{-1+3i)x / (-1 + 3i)
= Im (-1 - 3i) e^{1+3i)x / (-1 + 3i)(-1 - 3i)
= Im (-1- 3i) e^-x (cos3x + i sin 3x) / 10
= e^-x(-sin 3x - 3 cos 3x) / 10

## Answers and Replies

Cyosis
Homework Helper
Haven't looked at q2 and q3 yet, but the argument you calculated for q1 is wrong. Draw z in the complex plane and calculate the angle using normal trigonometry. As a result the rest of a and b is also wrong, although you used the correct method.

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