Can someone show me how to do this step by step?

Question

If p = the sq. root of 3 + i, and q = -1 -i, find pq and p/q in cartesian form. Also find the modulus and argument of p
thankyou

look just don't answer if you don't understand the question. It's important to me, that's all you need to know

Answer 1

For easier understanding I am using "v3" instead of sq.root of 3.

p = v3 + i
q = -1 - i

1. pq = (v3 + i) (-1 - i)
= -v3 - v3i - i + 1
= 1-v3 + i (-1-v3)
This is in the form x +iy for which the cartesian form is (x,y)

Similarly pq = (1-v3,-1-v3)

2. p/q = (v3 + i) / (-1 - i)
rationalise the denominator by multiplying with (-1+i) in numerator and denominator.

=(v3i - v3 - 1 - i) / 2 As (a+ib)(a-ib) = a^2 + b^2
= (-1 - v3)/2 + i(v3 - 1)/2

Finding the coordinates as in the form of pq.
( -1/2 - v3/2 , v3/2 - 1/2 )

3. If z = a + ib then the modulus of z = Sq.Rt. of (a^2 + b^2)
modulus of z is written as IzI.

Therefore IpI = SqRt. (3 + 1) = Sq.Rt. of 4 = 2
Hence IpI = 2.

4. Argument of z = a + ib is Tan^(-1) {b/a}
It is read as Tan inverse or Arc Tan b/a.

For p,
Arg. p = Arc. Tan (1/v3)
Hence Arg (p) = Pie / 6 radians (or) 30 degrees.

The four answers are as follows :
pq = (1-v3,-1-v3).
p/q = ( -1/2 - v3/2 , v3/2 - 1/2 ).
IpI = 2.
Arg (p) = Pie / 6 radians (or) 30 degrees.

Hope you had understood the explanation clearly.

Answer 2

when working with complex notation, you have to think in 2 dimensions... the complex plane.

if p=sqrt(3)+i, then modulus is (pytagore) sqrt(sqrt(3)²+1²)=sqrt(3+1)=2
argument, would be alpha in cos(alpha)=sqrt(3)/2
(you know this one, don't you? you should know this alpha by heart.)
draw it, you'll see why.

p*q=(sqrt(3)+i)*(-1-i) this one is easy, you only have to go fo it, and remember not to mix i's and non i's (and that i²=-1)
p*q= sqrt(3)*(-1) + sqrt(3)*(-i) + (-1)i - i²
=-sqrt(3)+1 -(sqrt(3)+1)i

p/q is more difficult.
well it's easier to multiply and divide complex numbers in modulus/argument notation, but with this notation, you can still do it if you remeber to use (a-b)(a+b)=a²-b²

p/q=-(sqrt(3)+i)/(1+i)
p/q=-(sqrt(3)+i)*(1-i)/(1-i²)
=-(sqrt(3)+i)*(1-i)/2.
i'll let you finish from here. you see how it's done?

Answer 3

pq = (V(3) + i)(-1-i) = -V3 -iV3 +1 (because -i*i = 1)

guess p/q = a + bi
p/q is (V(3) + i) / (-1-i). So
a + bi = (V(3) + i) / (-1-i). So
(a + bi)(-1-i) = (V(3) + i) or
(-a +b) - (a+b)i = V(3) + i.
This is true if
-a + b = V3 and a+b = -1
The solution is a = -(1 + V3)/2 and b = (-1 + V3)/2
So p/q = -(1 + V3)/2 + (-1 + V3)i/2

Answer 4

since sq. root of (a+ib)=[sq.root of{sq.root of(a^2+b^2)+a)}/2] +[sq.root of{sq.root of(a^2+b^2) - a)}/2]
property in complex numbers i^2 = -1.
here a=3, b=1 then sq. root of(a^2+b^2)=sq.root of 10=3.16
so, p=sq.root of {(3.16+3)/2}+i[sq.root of {(3.16 - 3)/2}]
then p=sq.root of {(4.58}+i[sq.root of {(0.08}]
p=(2.14)+i(0.28)

modulus of p = sq. root of(2.14^2+0.28^2)=4.58+0.07=4.65
augument of p=tan inverse of (0.28/2.14)=tan inverse of(0.13)
given by q= -1-i;then 1/q=1/(-1-i) = (-1+i)/{(-1-i)(-1+i)}= (-1+i)/2
now, pq=(2.14+i(0.28))(-1-i)
pq= (-2.14+0.28)+i(-2.14-0.28)= -(1.86)-i(2.42)

now,p/q ={(2.14)+i(0.28)}(-1+i)/2= {(-2.14-0.28)+i(2.14-0.28)}/2
hence, p/q={-2.42+i(1.86)}/2 = (-1.21)+i(0.93)

Answer 5

oh man i always hated complex numbers. That's why i don't remember anything about them.

Answer 6

What the %$^#?!?!