Given z = 2 + i, u = 3 - 2i and v = -1/2 - (root 3)/2 i, evaluate z^2 - 3u^3 + 2v and give the answer in the form a + bi.

Step-by-step solution

1. 1

   Square z. Use $(a+b)^2=a^2+2ab+b^2$ with $i^{2}=-1$:

   $z^{2}=(2+i)^{2}=4+4i+i^{2}=4+4i-1=3+4i.$

2. 2

   Cube u in two stages. First square it:

   $u^{2}=(3-2i)^{2}=9-12i+4i^{2}=9-12i-4=5-12i.$

   Then multiply by $u$ again:

   $u^{3}=(5-12i)(3-2i)=15-10i-36i+24i^{2}=15-46i-24=-9-46i.$

   Squaring then multiplying is far less error-prone than expanding a binomial cube in one go.

3. 3

   Multiply by the coefficients. Distribute the $-3$ across both parts:

   $-3u^{3}=-3(-9-46i)=27+138i.$

   And

   $2v=2\left(-\frac12-\frac{\sqrt3}{2}i\right)=-1-\sqrt{3}\,i.$

4. 4

   Add the real parts.

   $3+27-1=29.$

5. 5

   Add the imaginary parts. These do not merge into a single decimal, since one carries a surd:

   $4+138-\sqrt{3}=142-\sqrt{3}.$

6. 6

   Write the answer and check its size.

   $z^{2}-3u^{3}+2v=29+\left(142-\sqrt{3}\right)i\approx 29+140.268\,i.$

   The $u^3$ term dominates, as expected: $|u|=\sqrt{13}$, so $|u^{3}|=13\sqrt{13}\approx 46.9$ and $|-3u^3|\approx 140.6$, far larger than $|z^2|=5$ or $|2v|=2$. Incidentally $v$ has modulus $1$ and argument $-\tfrac{2\pi}{3}$, so it is a primitive cube root of unity.