Solve {left(frac{-9+9sqrt{3}texttt{i}}{2}right)}^2

$${ \left( \frac{ -9+9 \sqrt{ 3 } \texttt{i} }{ 2 } \right) }^{ 2 }$$

Evaluate

$$\frac{-81\sqrt{3}i-81}{2}\approx -40.5-70.148057707i$$

Solution Steps

Multiply 9 and i to get 9i.

$$\left(\frac{-9+9i\sqrt{3}}{2}\right)^{2}$$

To raise \frac{-9+9i\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(-9+9i\sqrt{3}\right)^{2}}{2^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-9+9i\sqrt{3}\right)^{2}.

$$\frac{81-162i\sqrt{3}-81\left(\sqrt{3}\right)^{2}}{2^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{81-162i\sqrt{3}-81\times 3}{2^{2}}$$

Multiply -81 and 3 to get -243.

$$\frac{81-162i\sqrt{3}-243}{2^{2}}$$

Subtract 243 from 81 to get -162.

$$\frac{-162-162i\sqrt{3}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{-162-162i\sqrt{3}}{4}$$

Expand

$$\frac{-81\sqrt{3}i-81}{2}$$

Solution Steps

Multiply 9 and i to get 9i.

$$\left(\frac{-9+9i\sqrt{3}}{2}\right)^{2}$$

To raise \frac{-9+9i\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(-9+9i\sqrt{3}\right)^{2}}{2^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-9+9i\sqrt{3}\right)^{2}.

$$\frac{81-162i\sqrt{3}-81\left(\sqrt{3}\right)^{2}}{2^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{81-162i\sqrt{3}-81\times 3}{2^{2}}$$

Multiply -81 and 3 to get -243.

$$\frac{81-162i\sqrt{3}-243}{2^{2}}$$

Subtract 243 from 81 to get -162.

$$\frac{-162-162i\sqrt{3}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{-162-162i\sqrt{3}}{4}$$