$\sqrt{ 3 }-2i-i \times \left( 2-i \times \left( \sqrt{ 3 }+4 \right) \right)$
Distribute $-i$ through the parentheses$\sqrt{ 3 }-2i-2i+i \times i \times \left( \sqrt{ 3 }+4 \right)$
The factor $i$ repeats $2$ times, so the base is $i$ and the exponent is $2$$\sqrt{ 3 }-2i-2i+{i}^{2} \times \left( \sqrt{ 3 }+4 \right)$
Distribute ${i}^{2}$ through the parentheses$\sqrt{ 3 }-2i-2i+{i}^{2}\sqrt{ 3 }+4{i}^{2}$
By definition ${i}^{2}=-1$$\sqrt{ 3 }-2i-2i-1\sqrt{ 3 }+4{i}^{2}$
By definition ${i}^{2}=-1$$\sqrt{ 3 }-2i-2i-1\sqrt{ 3 }+4 \times \left( -1 \right)$
Any expression multiplied by $1$ remains the same$\sqrt{ 3 }-2i-2i-\sqrt{ 3 }+4 \times \left( -1 \right)$
Any expression multiplied by $-1$ equals its opposite$\sqrt{ 3 }-2i-2i-\sqrt{ 3 }-4$
Since two opposites add up to $0$, remove them from the expression$-2i-2i-4$
Collect like terms$-4i-4$
Use the commutative property to reorder the terms$-4-4i$
To find the conjugate of a complex number $a+bi$, change the sign of the imaginary part$-4+4i$