﻿ If z_1,z_2\in C such that 2z^2_1+z^2_2=2z_1z_2 and z_1+z_2 is pure imaginary number, show that the

#### Question

If z_1,z_2\in C such that 2z^2_1+z^2_2=2z_1z_2 and z_1+z_2 is pure imaginary number, show that the complex number expression 3z_1-2z_2 is real number.

Collected in the board: Imaginary number

Steven Zheng posted 1 week ago

Since z_1+z_2 is pure imaginary number, let z_1= a+bi, then z_2 = -a+ci (a,b,c\in R).

Substitute z_1,z_2 to the equation 2z^2_1+z^2_2=2z_1z_2

2(a+bi)^2+(-a+ci)^2=2(a+bi)(-a+ci)

Expand brackets

2a^2-2b^2+4abi+a^2-c^2-2aci=-2a^2+2aci-2abi-2bc

Rearrange by real and imaginary parts for both sides

3a^2-2b^2-c^2+(4ab-2ac)i = -2a^2-2bc+(2ac-2ab)i

Comparing the real and imaginary parts gives

3a^2-2b^2-c^2= -2a^2-2bc
(1)
4ab-2ac=2ac-2ab
(2)

Solving the equation (2)

3ab=2ac\implies a(3b-2c)=0

Then,

a=0 or 3b=2c

Case 1,if a=0

Equation (1) becomes

-2b^2-c^2=-2bc

b^2+(b-c)^2=0

Then we get b=0 or b=c.

If b=0, then c=0. 3z_1-2z_2=0, which is real number.

If b=c, then 2z^2_1+z^2_2=2z_1z_2 will not be true. So cancel the case a=0.

Case 2, if 3b=2c (a\ne=0) , then b=\dfrac{2}{3}c.

3z_1-2z_2 = 3(a+bi)-2(-a+ci)

=5a+(3b-2c)i

=5a

Therefore, 3z_1-2z_2 is real number

Steven Zheng posted 1 week ago

Since z_1+z_2 is pure imaginary number, let z_1+z_2=ki (k\in R)

From the given condition 2z^2_1+z^2_2=2z_1z_2

Multiply 5 on both sides

10z^2_1+5z^2_2=10z_1z_2

Subtract z^2_1+z^2_2 from both sides

9z^2_1+4z^2_2=-z^2_1-z^2_2+10z_1z_2

Subtract 6z_1z_2 from both sides

9z^2_1+4z^2_2-6z_1z_2=-z^2_1-z^2_2+4z_1z_2

Apply perfect square identity

(3z_1-2z_2)^2=-(z_1+z_2)^2

Taking square root on both sides gives

3z_1-2z_2 = \pm i(z_1+z_2)=\pm i\cdot ki = \pm k

which is a real number.

Steven Zheng posted 1 week ago

Scroll to Top