Solve (4-isqrt{2})(4+isqrt{2})

$$( 4 - i \sqrt { 2 } ) ( 4 + i \sqrt { 2 } )$$

Evaluate

$$18$$

Solution Steps

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Calculate 4 to the power of 2 and get 16.

$$16-\left(i\sqrt{2}\right)^{2}$$

Expand \left(i\sqrt{2}\right)^{2}.

$$16-i^{2}\left(\sqrt{2}\right)^{2}$$

Calculate i to the power of 2 and get -1.

$$16-\left(-\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{2} is 2.

$$16-\left(-2\right)$$

The opposite of -2 is 2.

$$16+2$$

Add 16 and 2 to get 18.

$$18$$

Real Part

$$18$$

Solution Steps

Consider \left(4-i\sqrt{2}\right)\left(4+i\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$Re(4^{2}-\left(i\sqrt{2}\right)^{2})$$

Calculate 4 to the power of 2 and get 16.

$$Re(16-\left(i\sqrt{2}\right)^{2})$$

Expand \left(i\sqrt{2}\right)^{2}.

$$Re(16-i^{2}\left(\sqrt{2}\right)^{2})$$

Calculate i to the power of 2 and get -1.

$$Re(16-\left(-\left(\sqrt{2}\right)^{2}\right))$$

The square of \sqrt{2} is 2.

$$Re(16-\left(-2\right))$$

The opposite of -2 is 2.

$$Re(16+2)$$

Add 16 and 2 to get 18.

$$Re(18)$$

The real part of 18 is 18.

$$18$$