Porisma

Worked solution

Reuse a cube after multiplying by i

Without using a calculator, express (2−i)3(2-i)^3 in the form a+iba+ib.

Hence, find a root of (z−i)3=−11−2i(z-i)^3=-11-2i.

CJC · Example 13

Express the cube in algebraic form

Answer

(2−i)3=2−11i(2-i)^3=2-11i.

Expand the square first, then multiply:

(2−i)3=(4−4i+i2)(2−i)=(3−4i)(2−i)=6−11i+4i2=2−11i.\begin{aligned}(2-i)^3&=(4-4i+i^2)(2-i)\\&=(3-4i)(2-i)\\&=6-11i+4i^2\\&=2-11i.\end{aligned}

How did this part go?

Use the earlier cube to find one root

Answer

z=1+3iz=1+3i is a root.

Recall −i=i3-i=i^3. Reuse the first result:

\end{aligned}$$ Choosing $z-i=i(2-i)$ therefore supplies one root: $$z=i(2-i)+i=1+3i.$$ This finds **a** root, as requested. Equal complex cubes do not force equal bases; no claim that this is the only root is needed.
How did this part go?