pls help!!

Let z=x+yi, where x and y are real numbers. Then $$\begin{align*} (3 + 4i)z^3 &= (3 + 4i)(x^3 + 3ix^2 y - 3xy^2 - y^3) \ &= 3x^3 + 12ix^2 y + 12xy^2 - 3y^3 + 4x^2 + 4xy - 4y^2 i. \end{align*}$$ Also, z5=x5+5x4y−10x3y2+10x2y3−5xy4+y5, so $$\begin{align*} |(3 + 4i)z^3 - z^5|^2 &= (3x^3 + 12ix^2 y + 12xy^2 - 3y^3 + 4x^2 + 4xy - 4y^2 i)^2 \ &\quad - (x^5 + 5x^4 y - 10x^3 y^2 + 10x^2 y^3 - 5xy^4 + y^5)^2 \ &= 9x^6 + 48x^4 y^2 + 36x^2 y^4 + 9y^6 - 72x^5 y + 144x^3 y^3 - 72x^2 y^5 + 36xy^7 \ &\quad - x^{10} - 50x^8 y + 100x^6 y^2 - 100x^4 y^4 + 50x^2 y^6 - y^{10} \ &= -x^{10} + 9x^6 - 50x^8 y + 144x^6 y^2 - 72x^4 y^4 + 72x^2 y^5 - 9y^6 + y^{10} \ &= (x^6 - 10x^4 y + 36x^2 y^2 - 36xy^4 + y^6)^2 - (x^{10} - 9x^6 + y^{10})^2. \end{align*}$$ By AM-GM, $$\begin{align*} x^6 - 10x^4 y + 36x^2 y^2 - 36xy^4 + y^6 &\ge 5 \sqrt[5]{(x^6)(x^4 y)(x^2 y^2)(xy^4)(y^6)} \ &= 5x^2 y^2, \end{align*}$$ and $$\begin{align*} x^{10} - 9x^6 + y^{10} &\ge 5 \sqrt[5]{(x^{10})(9x^6)(y^{10})} \ &= 45x^6 y^{10}. \end{align*}$$ Therefore, $$\begin{align*} |(3 + 4i)z^3 - z^5|^2 &\ge (5x^2 y^2)^2 - (45x^6 y^{10})^2 \ &= 25x^4 y^4 - 2025x^{12} y^{12} \ &= -2000x^{12} y^{12} + 25x^4 y^4 \ &= (-2000x^6 y^6 + 25)^2. \end{align*}$$ Since x and y are real, [(-2000x^6 y^6 + 25)^2 \ge 25^2 = 625,]so ∣(3+4i)z3−z5∣2≥625. Hence, ∣(3+4i)z3−z5∣≥25.

Equality occurs when x=1 and y=0, so the maximum distance is 25​.