Solve i+5:1-6|-sqrt{9}+(1-sqrt{2})^0-(-3)

$$i + 5 : 1 - 6 | - \sqrt { 9 } + ( 1 - \sqrt { 2 } ) ^ { 0 } - ( - 3 )$$

Evaluate

$$-1+i$$

Solution Steps

Anything divided by one gives itself.

$$i+5-6|-\sqrt{9}+\left(1-\sqrt{2}\right)^{0}-\left(-3\right)|$$

Calculate the square root of 9 and get 3.

$$i+5-6|-3+\left(1-\sqrt{2}\right)^{0}-\left(-3\right)|$$

Calculate 1-\sqrt{2} to the power of 0 and get 1.

$$i+5-6|-3+1-\left(-3\right)|$$

Add -3 and 1 to get -2.

$$i+5-6|-2-\left(-3\right)|$$

The opposite of -3 is 3.

$$i+5-6|-2+3|$$

Add -2 and 3 to get 1.

$$i+5-6|1|$$

The modulus of a complex number a+bi is \sqrt{a^{2}+b^{2}}. The modulus of 1 is 1.

$$i+5-6\times 1$$

Multiply 6 and 1 to get 6.

$$i+5-6$$

Do the additions.

$$-1+i$$

Real Part

$$-1$$

Solution Steps

Anything divided by one gives itself.

$$Re(i+5-6|-\sqrt{9}+\left(1-\sqrt{2}\right)^{0}-\left(-3\right)|)$$

Calculate the square root of 9 and get 3.

$$Re(i+5-6|-3+\left(1-\sqrt{2}\right)^{0}-\left(-3\right)|)$$

Calculate 1-\sqrt{2} to the power of 0 and get 1.

$$Re(i+5-6|-3+1-\left(-3\right)|)$$

Add -3 and 1 to get -2.

$$Re(i+5-6|-2-\left(-3\right)|)$$

The opposite of -3 is 3.

$$Re(i+5-6|-2+3|)$$

Add -2 and 3 to get 1.

$$Re(i+5-6|1|)$$

The modulus of a complex number a+bi is \sqrt{a^{2}+b^{2}}. The modulus of 1 is 1.

$$Re(i+5-6\times 1)$$

Multiply 6 and 1 to get 6.

$$Re(i+5-6)$$

Do the additions in i+5-6.

$$Re(-1+i)$$

The real part of -1+i is -1.

$$-1$$