Solve for: (sqrt(3)-2i)-i * (2-i * (sqrt(3)+4))

Expression: $\left( \sqrt{ 3 }-2i \right)-i \times \left( 2-i \times \left( \sqrt{ 3 }+4 \right) \right)$

Remove unnecessary parentheses

$\sqrt{ 3 }-2i-i \times \left( 2-i \times \left( \sqrt{ 3 }+4 \right) \right)$

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

$-4+4i$