evaluating complex numbers?

Question

evaluate (1/1+i) + (1/2+i) + (1/3+1)
and i have to write the answer in the form a + bi (with a, b both being real numbers)
and simplify ((1/sqrt2) + (1/sqrt2i))^4

i'm all mathed out.......if you can please help it would be heaps appreciated

the first evaluate should read
(1/1+i) + (1/2+i) + (1/3+i)
NOT (1/3+1)

Answer 1

[1/(1+i)]+ [1/(2+i)]+ [1/(3+i)]

multiply and divide by conjugate of each term's denominator
in fact if (a+ib) is complex no then its conjugate is (a - ib) and their product is (a^2+b^2) as i^2 = -1

[(1-i)/(1+i)(1- i)]+ [(2 - i)/(2+i)(2-i)]+ [(3 - i)/(3+i)(3 - i)]

[(1-i)/(1+1)]+ [(2 - i)/(4+1)]+ [(3 - i)/(9+1)]

[(1-i)/(2)]+ [(2 - i)/(5)]+ [(3 - i)/(10)] collect real & imag. parts

[(1/2)+ (2/5)+(3/10)] - i [(1/2)+ (1/5)+(1/10)]

[(5+4+3)/10] - i [(5+2+1) /10]

[12/10] - i [8/10]

[6/5] - i [4/5]

(a+ib) form a = 6/5, b = - (4/5)
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simplify ((1/sqrt2) + (1/sqrt2i))^4

[(1/sqrt2) + (1/sqrt2i)]^4

(1/sqrt2)^4 * [1+ (1/ i)]^4

multiply and divide second term by ( i)

(1/sqrt2)^4 * [1+ i (1/ -1)]^4 (i^2 = -1)
(1/sqrt2)^3 * [(1/sqrt2) - i * (1/sqrt2)]^4
now cos(pi/4)= (1/sqrt2) = sin (pi/4)
(1/sqrt2)^3 * [cos(pi/4) - i * sin(pi/4)]^4

now [cos(x) - i * sin(x)]^n = [cos(nx) - i * sin(nx)]

(1/sqrt2)^3 * [cos(pi) - i * sin(pi)]
(1/sqrt2)^3 * [ -1 - i * 0]
= - (1/sqrt2)^3 = - (1/(2)^1/2)^3
= - (1/(2)^3/2) = - [1 / 2 sqrt2]

Answer 2

Common Denominator is (1 + i).(2 + i).(3 + i)
= (1 + 3i) (3 + i) = 10i

and Numerator becomes:-
(2 + i).(3 + i) + (1 + i).(3 + i) + (1 + i).(2 + i)
=(5 + 5 i) + (2 + 4 i) + (1 + 3 i)
= 8 + 12 i

Now have (8 + 12i) / 10i

= -10i (8 + 2i) / 100
= (- 80 i + 20) / 100
= (20/100) ( 1 - 4i)
= (1/5) (1 - 4i)

Question 2
1/√2 + (1 / √2i)^4
= 1 / √2 + 1 / 4
= (1 / 4).[(2.√2) + 1]

Answer 3

1/1+1/2+1/3 + 3i

6/6 + 3/6 + 2/6 + 3i

11/6 + 3i <<< answer

sqrt2/2 + 1/4i <<< not sure

Answer 4

add the a's...1/1+1/2+1/3=
then the b's... i + i + i =
does that get ya closer?
:)