Answer to Question #124550 in Algebra for jim

Question

3. (a) Show that, for any complex number z, zz = |z|^2., z + z = 2Re(z) and Re(z) ≤ |z|. Hence
show that
i. |z1 + z2|^2 = |z1|^2 + |z2|^2 + 2Re(z1z2),
ii. |z1 + z2| ≤ |z1| + |z2|,
where Re(z) is the real part of z and *z the conjugate of z.

(b) If z1 = 1 + 2i, find the set of values of z2 for which
(i) |z1 + z2| = |z1| + |z2|
(ii) |z1 + z2| = |z1| − |z2|.

Answer 1

z=a+bi

can write a=Re(z)

if z+z=2Re(z) z=a b=0

∣ z∣² =a²

z₁= a₁ z₂=a₂

z₁+z₂= a₁+a₂

∣z1+z2 ∣²=a₁²+a₂²+2a₁* a₂

∣ z₁∣ =a₁

∣z₂∣ =a2

Re(a₁a₂) =a₁a₂

∣ z₁+z₂∣² =∣z₁∣² +∣ z₂∣ ²+2Re(a₁a₂)

property of the modulus and argument of a complex number

  |z₁|−|z₂|≤|z₁+z₂|≤|z₁|+|z₂|

(b)

(i) z₁=1+2i let be z₂=a+bi

∣ z₁+z₂∣ =((1+a)²+(2+b)²)^1/2

∣ z₁∣ =(1²+2²)^1/2=5^1/2 

∣ z₂∣ =(a²+b²)^1/2

((1+a)² +(2+b)²)^1/2=5 ^1/2+(a²+b²)^1/2

square everything

1+2a+a²+4+4b+b²=5+2(5)^1/2(a²+b²)^1/2+a²+b²

a+2b=5^1/2(a²+b²)^1/2

square everything

a²+4ab+4b²=5(a²+b²)

4a²-4ab+b²=0

2a=b where both numbers are positive

(ii)((1+a)²+(2+b)²)^1/2=5^1/2-(a²+b²)^1/2

1+2a+a²+4+4b+b²=5-2(5)^1/2(a²+b²)^1/2+a²+b²

2a=b

where both numbers are negative