I need to find the maximum value of $|iz+3-4i|$ given that $|z|\leq4$
$|iz+3-4i|\leq |iz|+|3-4i|$
$|iz+3-4i|\leq 4+5$
But I could write an inequality for $|3-4i|$ whose maximum value would be 7, not 5 so I get the maximum value to be 11 when doing so rather than the original answer of 9.
What went wrong here? , thanks in advance.
For a point in a disk of radius $4$ centered at the origin, what is the maximum distance of that point from $(3,-4)$? Clearly, the farthest (and closest!) point away is the one that is on the line that joins $(3,-4)$ and the circle center $(0,0)$. Since the distance between $z$ and the center is at most $4$, and the distance from the center to $(3,-4)$ is $5$, the maximum distance is $4+5 = 9$ and the minimum is $5-4 = 1$. The fact that $z$ is multiplied by $i$ is of no consequence, because we can set $iz = w$ and then $|z| \le 4$ is equivalent to $|w/i| = |w| \le 4$.
Nothing seems to have gone wrong. $9 = 4 + 5 \leq |iz| + |3 - 4i| \leq |iz| + |3| + |4i| \leq 4 + 3 + 4 = 11$. And, well, $9 < 11$.
A minor simplification is $|iz+3-4i|=|i(z-4-3i)|=|z-4-3i|$, which dispenses of the distracting factor of $i$. Moreover, it allows us to interpret $|z-4-3i|$ as the distance between $z$ and $4+3i$. That suggests a tidy geometric interpretation and a simple answer.
The reason is $|3-4i|$ represents a real number associated to the Eulidean norm of the point which you call the distance from origin. There is no way you can maximize a real number $c$(say) unless it is a real variable.
Put $z = a + bi\implies |iz+3-4i|= |i(a+bi)+3-4i|=|3-b+(a-4)i|=\sqrt{(a-4)^2+(b-3)^2}$. The problem translates to: Maximize $f(a,b) = \sqrt{(a-4)^2+(b-3)^2}$ given that $a^2+b^2 \le 16$. First observe that if the max occurs at an interior point inside the disk $D: a^2+b^2 \le 16$ at $M$, then consider the point of intersection $B$ of the line joining $A = (4,3)$ to $M$ and the circle $C: a^2+b^2 = 16$, we have: $AM \le AB$ and $AB$ is the max value. This can be proven rigorously but is quite easy. Thus the constraint is reduced to the circle $a^2+b^2 = 16$, and $f(a,b) = \sqrt{a^2-8a+16+b^2-6b+9}=\sqrt{41-8a-6b}$. Again observe that $f(a,b)$ is maximized iff $g(a,b) = 41 - 8a-6b$ is maximized under the condition $a^2+b^2=16$. This can be done with Lagrange Multiplier from calculus. Indeed, $\nabla g(a,b) = \lambda\nabla h(a,b), h(a,b) = a^2+b^2-16\implies (-8,-6)=(2\lambda a, 2\lambda b)\implies 2\lambda a= -8, 2\lambda b = -6\implies \lambda a = -4, \lambda b = -3\implies \lambda^2(a^2+b^2)= (-4)^2+(-3)^2=25\implies 16\lambda^2=25\implies \lambda = \pm \dfrac{5}{4}$. We can see that only $\lambda = \dfrac{5}{4}$ yields the max value, and it corresponds to $a= -\dfrac{16}{5}, b = -\dfrac{12}{5}\implies g_{\text{max}} = 41-8\cdot \dfrac{-16}{5}-6\cdot \dfrac{-12}{5}=81\implies f_{\text{max}}= \sqrt{g_{\text{max}}}=\sqrt{81} = 9$, and this is the maximum value of $|iz+3-4i|$.