Answer to Question #124550 in Algebra for jim

Question #124550
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|.
1
Expert's answer
2020-07-02T19:22:03-0400

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+z22 =∣z12 +∣ 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=51/2 

∣ 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=51/2(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=51/2-(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



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS