FOIL the product out:

\(\displaystyle \left (7 - 2i \right )\left (7 - 3i \right )\)

\(\displaystyle = 7 \cdot 7 - 7 \cdot 3i - 7 \cdot 2i + 2i \cdot 3i\)

\(\displaystyle = 49 -21i - 14i + 6 i^{2}\)

\(\displaystyle = 49 -21i - 14i + 6 (-1)\)

\(\displaystyle = 49 -6 -21i - 14i\)

\(\displaystyle = 43 -35i\)

Use the square of a binomial pattern to multiply this:

\(\displaystyle \left ( 8 - 3i\right ) ^{2}\)

\(\displaystyle = 8 ^{2}- 2 \cdot8\cdot3i + \left ( 3i\right )^{2}\)

\(\displaystyle = 64- 48i + 3^{2}i^{2}\)

\(\displaystyle = 64- 48i + 9 (-1)\)

\(\displaystyle = 64-9 - 48i\)

\(\displaystyle = 55 - 48i\)

This is a product of an imaginary number and its complex conjugate, so it can be evaluated using this formula:

\(\displaystyle \left (a + bi \right )\left (a -bi \right ) = a^{2} + b ^{2}\)

\(\displaystyle \left (1.1 + 0.6i \right )\left (1.1 -0.6 i \right ) = 1.1^{2} + 0.6 ^{2} = 1.21 + 0.36 = 1.57\)

This is a product of an imaginary number and its complex conjugate, so it can be evaluated using this formula:

\(\displaystyle \left (a - bi \right )\left (a +bi \right ) = a^{2} + b ^{2}\)

\(\displaystyle \left (5 - 3i \right )\left (5 +3i \right ) = 5 ^{2} + 3 ^{2}= 25 + 9 = 34\)

\(\displaystyle a*b = a ^{5}-b^{3}\)

\(\displaystyle \left (2 i \right )* \left (3i \right ) = \left (2 i \right ) ^{5}-\left (3i \right ) ^{3}\)

\(\displaystyle = 2^{5} i^{5} - 3^{3} i ^{3}\)

\(\displaystyle = 32\cdot i^{4} \cdot i - 27 \cdot (-i )\)

\(\displaystyle = 32\cdot 1 \cdot i +27 i\)

\(\displaystyle = 32 i +27 i\)

\(\displaystyle = 59 i\)

\(\displaystyle a*b = a+3ib\)

\(\displaystyle (4+7i) * (3-8i) = (4+7i) + 3i (3-8i)\)

\(\displaystyle = 4+7i + 3i \cdot 3- 3i \cdot 8i\)

\(\displaystyle = 4+7i + 9i - 24i ^{2}\)

\(\displaystyle = 4+7i + 9i - 24 (-1)\)

\(\displaystyle = 4+7i + 9i- (-24)\)

\(\displaystyle = 4+7i + 9i +24\)

\(\displaystyle = 28 +16i\)

\(\displaystyle (3i)^{-5} = \frac{1}{(3i)^{5} }= \frac{1}{3^{5}i^{5} } = \frac{1}{243 i^{4} \cdot i } = \frac{1}{243 \cdot 1 \cdot i }\)

\(\displaystyle = \frac{1}{243 i } \cdot = \frac{1\cdot i}{243 i \cdot i } = \frac{i}{243 (-1)} = - \frac{1}{243} i\)

\(\displaystyle a \wedge b = a \div (b+ 2i) = \frac{a}{b+2i}\)

\(\displaystyle 8 \wedge 4 = \frac{8}{4+2i}\)

Multiply both numerator and denominator by the conjugate of the denominator, \(\displaystyle 4 - 2i\), to rationalize the denominator:

\(\displaystyle 8 \wedge 4 = \frac{8(4-2i)}{(4+2i)(4-2i)}\)

\(\displaystyle = \frac{8\cdot 4-8\cdot 2i}{4^{2}+2^{2}}\)

\(\displaystyle = \frac{32-16i}{16+4}\)

\(\displaystyle = \frac{32-16i}{20}\)

\(\displaystyle = \frac{32 }{20} - \frac{ 16}{20}i\)

\(\displaystyle = \frac{8}{5} -\frac{ 4}{5}i\)

\(\displaystyle \bigstar a = a^{2}i + a\)

\(\displaystyle \bigstar (7i) = (7i)^{2}i + 7i\)

\(\displaystyle = 7^{2}\cdot i^{2} \cdot i + 7i\)

\(\displaystyle = 49\cdot (-1) \cdot i + 7i\)

\(\displaystyle = -49 i + 7i\)

\(\displaystyle =-42i\)

If \(\displaystyle \bigstar N = 0\), then 

\(\displaystyle N^{2}i - 4 N= 0\). 

Distribute out \(\displaystyle N\) to yield

\(\displaystyle N (Ni - 4 )= 0\)

Either \(\displaystyle N = 0\) or \(\displaystyle N i - 4 = 0\). However, we are given that \(\displaystyle N \ne 0\), so 

\(\displaystyle N i - 4 = 0\)

\(\displaystyle N i = 4\)

\(\displaystyle \frac{N i }{i}= \frac{4}{i}\)

\(\displaystyle N= \frac{4}{i} = \frac{4 \cdot i}{i\cdot i}= \frac{4i}{-1} = -4i\)