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.

gb1falcon May 7, 2023

#1**0 **

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∣.

Guest May 7, 2023