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Given two complex numbers $z,w$ with unit modulus (i.e., $ |z|=|w|=1$), which of the following statements will always be correct?

a.) $|z+w|\lt\sqrt2$ and $|z-w|\lt\sqrt2$

b.) $|z+w|\le\sqrt2$ and $|z-w|\ge\sqrt2$

c.) $|z+w|\ge\sqrt2$ or $|z-w|\ge\sqrt2$

d.) $|z+w|\lt\sqrt2$ or $|z-w|\lt\sqrt2$

Source [ISI entrance examination]

It is a multiple choice question and only one option is correct.

My approach: As modulus of $z$ and $w$ is 1. Let $z=e^{i\alpha_1}$ and $w=e^{i\alpha_2}$. Now,

$$|z+w|= |e^{i\alpha_1}+e^{i\alpha_2}|$$

$$|z+w|=|2 cos(\frac{\alpha_1-\alpha_2}{2})e^{\frac{i(\alpha_1+\alpha_2)}{2}}|$$

$$|z+w|=2 |cos(\frac{\alpha_1-\alpha_2}{2})|$$

I don't know how to proceed after this. Any help would be appreciated.

tryst with freedom
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Broly-29
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  • @MartinR You are right, but these are the options given in the question. – Broly-29 Aug 27 '20 at 14:40
  • @user376343 yeah, by using triangle inequality we know that this is the range of $|z+w|$. But these are the same options as given in the original paper and the answer is given as option C. – Broly-29 Aug 27 '20 at 14:46
  • Could there be a typo and it is c.) $|z+w|\ge\sqrt2$ or $|z-w|\ge\sqrt2$ ? – Martin R Aug 27 '20 at 14:53
  • @MartinR Yeah the answer is given as C. maybe there is some typo, i don't know. – Broly-29 Aug 27 '20 at 15:00
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    @MartinR I have cross verified with the original question paper and edited the question. I think there is a printing mistake in my book. – Broly-29 Aug 27 '20 at 16:06

1 Answers1

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(C) holds as a consequence of the parallelogram law:

$$ |z+w|^2 + |z-w|^2 = 2(|z|^2+|w|^2) = 4 $$ so that at least one of $|z+w|^2$ or $|z-w|^2$ must be $\ge 2$.

I leave it to you to find counterexamples for (A), (B), and (D). These can easily be found by choosing $z, w$ from $-1, 1, i$.

Martin R
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  • Thank you for correcting me – tryst with freedom Jun 14 '21 at 11:56
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    @Buraian: No problem! I don't think that this can be solved with the triangle inequality alone, because it becomes wrong with other norms which are not induced by an inner product. Try it with $\Vert x+iy \Vert = (x^4+y^4)^{1/4}$. – Martin R Jun 14 '21 at 12:02
  • I am having difficulty understanding this part of your comment : "because it becomes wrong with other norms which are not induced by an inner product.". Could you explain in a bit more detail? If I understood correctly, those concepts you mentioned are from linear algebra, are those necessary for speaking of length of complex numbers?/ How do they help us here? – tryst with freedom Jun 14 '21 at 12:06
  • @Buraian: The parallelogram law does not hold for arbitrary norms, but only if the norm is related to an inner product. See https://en.wikipedia.org/wiki/Parallelogram_law#The_parallelogram_law_in_inner_product_spaces – Martin R Jun 14 '21 at 12:08