Solve (sqrt{2}-3i)^2-(3-i)*(3+i)+3sqrt{2}i=

$$( \sqrt { 2 } - 3 i ) ^ { 2 } - ( 3 - i ) \cdot ( 3 + i ) + 3 \sqrt { 2 } i =$$

Evaluate

$$-3\sqrt{2}i-17\approx -17-4.242640687i$$

Solution Steps

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

$$\left(\sqrt{2}\right)^{2}-6i\sqrt{2}-9-\left(3-i\right)\left(3+i\right)+3i\sqrt{2}$$

The square of \sqrt{2} is 2.

$$2-6i\sqrt{2}-9-\left(3-i\right)\left(3+i\right)+3i\sqrt{2}$$

Subtract 9 from 2 to get -7.

$$-7-6i\sqrt{2}-\left(3-i\right)\left(3+i\right)+3i\sqrt{2}$$

Multiply 3-i and 3+i to get 10.

$$-7-6i\sqrt{2}-10+3i\sqrt{2}$$

Subtract 10 from -7 to get -17.

$$-17-6i\sqrt{2}+3i\sqrt{2}$$

Combine -6i\sqrt{2} and 3i\sqrt{2} to get -3i\sqrt{2}.

$$-17-3i\sqrt{2}$$

Real Part

$$-17$$