Solve (sqrt{3}+isqrt{2})-(sqrt{3}-isqrt{8})

$$( \sqrt { 3 } + i \sqrt { 2 } ) - ( \sqrt { 3 } - i \sqrt { 8 } )$$

Evaluate

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

Solution Steps

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

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

Multiply -1 and 2i to get -2i.

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

To find the opposite of \sqrt{3}-2i\sqrt{2}, find the opposite of each term.

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

Combine \sqrt{3} and -\sqrt{3} to get 0.

$$i\sqrt{2}+2i\sqrt{2}$$

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

$$3i\sqrt{2}$$

Real Part

$$0$$