I need help ..plz!regarding math. complex numbers?

Question

I need help with this question. anyone? plz. first correct answer gets ten points. and plz explain..clearly..thanx..
(a) express the complex numbers 1+ i and 1-i in the form r(cos Θ + sin Θ) where r is the modulus and Θ the argument of the complex number. Find the modulus and argument of (1+i)^2 / (1-i)^2.

(b) Given that z1 = 2 + i and z2 =-2 +4i, find, in the form a+bi,the complex number z such that 1/z=1/z1+1/z2.
find the modulus and argument of z, giving your answer for arg z in the range -π < arg z< π , corrext to 3 decimal places.

Answer 1

Im not exactly sure what part a is asking... but if you want to write 1+i and 1-i in polar form then its
√2 cis ¼π and √2 cis 1¾π

To find (1+i)^2 / (1-i)^2 it is fairly easy with polar form:
(√2 cis ¼π)^2 / (√2 cis 1¾π)^2 ........ simple substitution
2 cis ½π / 2 cis 3½π ....... multiplication and division rules
cis -3π
cis 1π .... equivalence: Θ = &Theta + 2πk
-1 + 0i ..... rectangular form.
-1


Part B, I wont use polar

z1 = 2 + i
1/z1 = 1/(2+i)
(2-i)/(2+i)(2-i)
(2-i)/(4+1)
2/5 - 1/5i

z2 = -2 + 4i
1/z2 = 1/(-2+4i)
(-2-4i)/(-2+4i)(-2-4i)
(-2-4i)/(4+16)
-2/20 - 4/20i

1/z = 1/z1 + 1/z2 =
(2/5 - 1/5i) + (-2/20 - 4/20i)
(4/10 - 2/10i) + (-1/10 - 2/10i)
3/10 - 4/10i

z = 1/1/z = 1/(3/10 - 4/10i) =
(3/10 + 4/10i)/(3/10 - 4/10i)(3/10 + 4/10i)
(3/10 + 4/10i)/(9/100 + 16/100)
(3/10 + 4/10i)/(25/100)
12/10 + 20/10i
6/5 + 2i

Answer 2

Slight typo in (a) - it should be "in the form r(cos θ + i sin θ)".

1 + i = √2 (1/√2 + i/√2)
= √2 (cos π/4 + i sin π/4)

1 - i = √2 (1/√2 - i/√2)
= √2 (cos -π/4 + i sin -π/4)

(1 + i)^2 / (1 - i)^2
= 2 (cos π/2 + i sin π/2) / 2 (cos -π/2 + i sin -π/2))
= cos π + i sin π
(= -1)
so the modulus is 1 and the argument is π.
Check: (1+i)^2 = 1 + 2i + i^2 = 2i, (1-i)^2 = 1 - 2i + i^2 = -2i. So the quotient is -1, as before.

(b) 1/z = 1/z1 + 1/z2 = z2/z1z2 + z1/z1z2 = (z1+z2)/z1z2
So z = z1z2 / (z1+z2) = (2+i)(-2+4i) / (5i)
= -i (-4+6i+4i^2) / 5(-i^2)
= -i (-8 + 6i) / 5
= 6/5 + (8/5)i.

|z| = √((6/5)^2 + (8/5)^2) = 2/5 √(3^2 + 4^2) = 2.
arg z = arcsin (4/5) = 0.927 to 3 d.p.

Answer 3

One approach is basically to %. any complicated variety and divide 10 by it to get yet another complicated variety. yet enable's search for some thing with integers contained in the kind (a + bi)(a - bi) Multiplying those supplies a^2 + b^2, so that you're searching for 2 sq. numbers that sum to 10. that would want to nicely be a million and 9 (or 9 and a million; order doesn't remember). 10 = (a million + 3i)(a million-3i) or 10 = (3 + i)(3 - i)

Answer 4

(a)
|1 + i|
= √(1² + 1²)
= √2

arg (1 + i)
= tan-1 1
= π/4

Hence (1 + i) = √2(cos (π/4) + i sin (π/4)) (ANS)

|1 - i|
= |1 + i|
= √2

arg (1 - i)
= π - tan-1 1
= 3π/4

Hence (1 - i) = √2(cos (3π/4) + i sin (3π/4)) (ANS)

|(1 + i)² / (1 - i)²|
= |(1 + i)|² / |(1 - i)|²
= (√2 / √2)²
= 1 (ANS)

arg ((1 + i)² / (1 - i)²)
= 2 arg (1 + i) - 2 arg (1 - i)
= π/2 - 3π/2
= -π (ANS)

(b)
1/z = 1/z1 + 1/z2
1/z = (z1 + z2) / (z1z2)
z = (z1z2) / (z1 + z2)
z = (-8 + 6i) / 5i
z = ((-8 + 6i) / 5i) X (5i / 5i)
z = 6/5 + 8i/5 (ANS)

|z|
= √((6/5)² + (8/5)²)
= √4
= 2 (ANS)

arg (z)
= tan-1 ((6/5) / (8/5))
= 0.644 rad (3sf) (ANS)

Answer 5

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