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# Very Hard Complex Numbers

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Let z be a complex number such that $$|z-1| \le 3$$. Find all possible values of $$|iz + 3 - 5i|$$. Express the answer as an interval.

May 7, 2023

#1
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We can solve this problem using complex numbers.

Let z=x+yi, where x and y are real numbers. Then ∣z−1∣=∣x+yi−1∣=∣(x−1)+yi∣≤3, which implies that ∣x−1∣≤3 and ∣y∣≤3. Therefore, −2≤x≤4 and −3≤y≤3.

We have that

\begin{align*}

|iz + 3 - 5i| &= |i(x + yi) + 3 - 5i| \\ &= |(ix) + (iy) + 3 - 5i| \\ &= |-y + 3 - 5i| \\ &= \sqrt{y^2 + (3 - 5i)^2} \\ &= \sqrt{y^2 + 9 + 25 + 30i} \\ &= \sqrt{y^2 + 44 + 30i}. \end{align*}

Since −3≤y≤3, we have that 0≤y^2+44≤55, so 0≤sqrt(y^2+44)​≤sqrt(55)​. Therefore, [0,sqrt(55)​]​ is the interval of all possible values of ∣iz+3−5i∣.

May 7, 2023
#2
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I like that approach, but I don't think it is entirely correct. In this problem, $$|z-1|=|x+yi-1|=|(x-1)+yi|=\sqrt{(x-1)^2+y^2}$$, because $$|x+yi|$$ signifies the magnitude of the number, which is $$\sqrt{x^2+y^2}.$$ I think that you interpreted it as absolute value.

gb1falcon  May 9, 2023
#3
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What do you expect the answer to look like ?

Would you regard

$$|z-(5-2i)|\leq4,$$

for example as an interval ?

Guest May 10, 2023