no teacher! MATH HELP! thnak you?

Question

i have 200 q's to do this summer. and im stuck on these 5. i have no one to consult with except you bright and intelligent people. please help...
#1. a) evaluate (1 + i)^2, where i = Squareroot(-1)
b) prove, by mathematical induction, that (1 + i)^(4n) = - 4^n, where n € N
c) hence or otherwise, find (1+ i)^32

#3. the variables x,y,z satisfy the simultaneous equations....
x + 2y + z = k
2x + y + 4z = 6
x - 4y + 5z = 9
where k is a constant.
a) (I) show that these equations do NOT have a unique solution.
(II) find the value of k for which the equations are consistent (that is, they can be solved).
b) for this value of k, find the general solutions of these equations

#5. solve the equation l e^(2x) - ( 1/ (x+2) ) l = 2
#4. the ratio of the fifth term to the twelfth term of a sequence in an arithmetic progression is 6 / 13. If each term of this sequence is positive, and the product of the first terms and the third term is 32, find the sum of the first 100 terms

#2. a farmer owns a triangular field ABC. The side [AC] is 104m, the side [AB] is 65m and the angle between these two sides is 60 degrees.
a) calculate the length of the third side of the field
b) find the area of the field in the form p *squareroot*(3), where p is an integer

(same prob) Let D be a point on [BC] such that [AD] bisects the 60 degree angle. The farmer divides the field into two parts by constructing a straight fence [AD] of length x meters.
c) (I) show that the area of the smaller part is given by (65x) / 4 and find an expression for the area of the larger part
(II) hence, find the value of x in the form q *squareroot* (3), where q is an integer
d) prove that BD / DC = 5 / 8

Answer 1

I'll help with your first question. Maybe someone else will help with others (or you could decide to be more reasonable with your requests).

a) (1 + i)^2 = 1 + 2i + i^2 = 1 + 2i + (-1) = 2i.

b) Let P(n) be the statement (1 + i)^(4n) = (-4)^n. Now,
(1 + i)^4 = [(1 + i)^2]^2 = [2i]^2 = 4i^2 = -4; that shows that P(1) is true.

Now assume that P(k) = (1 + i)^(4k) = (-4)^k. We have
(1 + i)^(4*(k+1)) = (1 + i)^(4k+4) = [(1 + i)^(4k)]*(1 + i)^4 =
[(-4)^(4k)]*(-4), by using P(k) and P(1). In turn, this last expression is = (-4)^(4*(k+1)), which shows that p(k + 1) is true whenever p(k) is true. This completes the proof.

c) From (b) we know (1 + i)^32 = (1 + i)^(4*8) = (-4)^8 =
(16)^4 = 256^2 = 65 536.

Answer 2

#1 For (1+i)^2 you use the FOIL method. Multiply the first terms, the outside terms, the inside terms, and the last terms.

a. (1+i)(1+i)= 1+i+i+i^2 = 1+2i-1 (i^2 always equals -1) = 2i

b. (1+i)^4n = (1+i)^2(1+i)^2n= (2i)^n(2i)^n = -4^n

c. so if (1+i)^2 =2i then (1+i)^32 would be 16 times that so it would be 32i (I think that is correct, Its been a while since Ive had complex numbers.

Answer 3

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Answer 4

Methinks you ask too much.