Solve {left(frac{3-sqrt{103}i}{2}right)}^2

$${ \left( \frac{ 3- \sqrt{ 103 } i }{ 2 } \right) }^{ 2 }$$

Evaluate

$$\frac{-3\sqrt{103}i-47}{2}\approx -23.5-15.223337348i$$

Solution Steps

To raise \frac{3-i\sqrt{103}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(3-i\sqrt{103}\right)^{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(3-i\sqrt{103}\right)^{2}}{4}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3-i\sqrt{103}\right)^{2}.

$$\frac{9-6i\sqrt{103}-\left(\sqrt{103}\right)^{2}}{4}$$

The square of \sqrt{103} is 103.

$$\frac{9-6i\sqrt{103}-103}{4}$$

Subtract 103 from 9 to get -94.

$$\frac{-94-6i\sqrt{103}}{4}$$

Expand

$$\frac{-3\sqrt{103}i-47}{2}$$

Solution Steps

To raise \frac{3-i\sqrt{103}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(3-i\sqrt{103}\right)^{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(3-i\sqrt{103}\right)^{2}}{4}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3-i\sqrt{103}\right)^{2}.

$$\frac{9-6i\sqrt{103}-\left(\sqrt{103}\right)^{2}}{4}$$

The square of \sqrt{103} is 103.

$$\frac{9-6i\sqrt{103}-103}{4}$$

Subtract 103 from 9 to get -94.

$$\frac{-94-6i\sqrt{103}}{4}$$