(Complex <span title="number)lz1l=1,lz2l=2,lz3l=4">number)lz1l=1,lz2l=2,lz3l...</span> How to prove z1+z2+z3 is not 0?

Question

I can not solve it,what to do.please help me

Answer 1

Let z1 = a+bi, where a and b are real numbers and i^2 = -1.

a and b must not = 0, else we would not have a complex number.

Then |z1| = sqrt(a^2+b^2) = 1 --&gt; a^2+b^2 = 1

Similarly |z2| = sqrt(c^2+d^2) = 2 and |z3| = sqrt(e^2+f^2) = 4.

This means c^2+d^2 = 4 and e^2+f^2 = 16

So a^2+b^2+c^2+d^2+e^2+f^2 = 21

If z1+z2+z3 = a+c+e +(b+d+f)i = 0, then a+c+e = 0 and b+d+f = 0.

Then (a+c+e)^2 = a^2+c^2+e^2 +2(ac+ae+ce) = 0 and,

(b + d + f)^2 = b^2+d^2+f^2 + 2(bd + be+ bf) = 0

So a^2+b^2+c^2+d^2+e^2+f^2 + 2(ac+ae+ce + bd+be+ef)=0

Then 21+ 2(ac+ae+ce + bd+be+ef)=0

Then ac+ae+ce + bd+be+ef = - 10.5

But this is not possible.

Thus z1+z2+z3 cannot = 0

Answer 2

The magnitude of a complex number is graphically its hypotenuse.

Since you know all of these and you are trying to prove that they can&#039;t sum to zero, these lines shouldn&#039;t be able to fit together in order to form a triangle. (i.e. You shouldn&#039;t be able to connect the vectors end to end and have them return to the origin.)

So use the law of cosines:

c² = a² + b² - 2·a·b·cos(θ)

Solve for the angle θ :

1² = 2² + 4² - 2·2·4·cos(θ)

1 = 4 + 16 - 16·cos(θ)

1 = 20 - 16·cos(θ)

-19 = -16·cos(θ)

19/16 = cos(θ)

Cosine outputs values between 1 and -1, but 19/16 is bigger than 1.

Therefore, no angle can form this triangle.

They cannot sum to 0.

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It should also be possible to observe that 4 - 2 - 1 = 1, so the other two lines cannot return to the origin.