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Suppose I want to find the maximum value of |a + 3+ 4i| where a is any complex number with |a| = 5. Now I assume b = 3+4i, resulting in the following inequality:

                                            |a + b| ≤ |a| + |b|

which gives the maximum value as 10.

But I considered another way of looking at this. Suppose we assume some complex number p= 3 + 0i and q = 0 + 4i. Then we get another inequality:

                                       |a + p + q| ≤ |a| + |p| + |q|

which when computed gives maximum value as 12.

Now as per my understanding, the correct answer is 10 because if we see consider this graphically, we get a and b as points on a circle of radius 5 units, whose maximum separation is equal to the diameter of the circle, that is, 10 units. But I am unable to clearly point out what is wrong in the second approach. I did not think there is any mistake in the second approach, yet it is tending towards an incorrect result. I would appreciate a well explained answer, clearly pointing out the errors.

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  • $\begingroup$ Hi and welcome to Math.SE. It would be preferable to use MathJax for mathematical expressions. You can get started here, and a more complete reference can be found here. $\endgroup$ Mar 29, 2021 at 8:14

2 Answers 2

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Your inequlities don't give the maximum value. [Please see also the comments by user3733558 below].

Let $c$ and $d$ be the real and imaginary parts of $a$. We want to maximize the square root of $(c+3)^{2}+(d+4)^{2}$. Expand the squares. Since $c^{2}+d^{2}=25$ it is enough to maximixe $6c+8d$. By Cauchy-Schwarz inequality the maximum value is attained when $c=6t$ and $d=8t$ for some $t$ and $t$ is determined by the condition $c^{2}+d^{2}=25$. I will let you finish.

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    $\begingroup$ If I may, I'd like to suggest a rephrasing for your first sentence: "Your inequalities don't give maximum values, they give upper bounds. Your second one remains valid, but is simply not as sharp as the first one." It seems important (at least to me) to dispel the OP's misunderstanding here. $\endgroup$ Mar 29, 2021 at 8:50
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$|a+b+c|\le|a|+|b|+|c|$ is proved by induction with the base case of $|a+b|\le|a|+|b|$, so we have $|a+b+c|\le|a|+|b+c|\le|a|+|b|+|c|$.

The inequality is generic, so in your case with $|p+q|\le|p|+|q|$, equality is when the arguments of $\vec p$ and $\vec q$ are the same.

You have used a specific $\vec p$ and $\vec q$, and thus equality is not reached, and there is an over-estimation.

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