Solve 1/4i-2i^2-i^3+i^4

$$\frac { 1 } { 4 i } - 2 i ^ { 2 } - i ^ { 3 } + i ^ { 4 }$$

Evaluate

$$3+\frac{3}{4}i=3+0.75i$$

Solution Steps

Multiply both numerator and denominator of \frac{1}{4i} by imaginary unit i.

$$\frac{i}{-4}-2i^{2}-i^{3}+i^{4}$$

Divide i by -4 to get -\frac{1}{4}i.

$$-\frac{1}{4}i-2i^{2}-i^{3}+i^{4}$$

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

$$-\frac{1}{4}i-2\left(-1\right)-i^{3}+i^{4}$$

Multiply 2 and -1 to get -2.

$$-\frac{1}{4}i-\left(-2\right)-i^{3}+i^{4}$$

The opposite of -2 is 2.

$$-\frac{1}{4}i+2-i^{3}+i^{4}$$

Calculate i to the power of 3 and get -i.

$$-\frac{1}{4}i+2-\left(-i\right)+i^{4}$$

The opposite of -i is i.

$$-\frac{1}{4}i+2+i+i^{4}$$

Do the additions.

$$i^{4}+2+\frac{3}{4}i$$

Calculate i to the power of 4 and get 1.

$$1+2+\frac{3}{4}i$$

Do the additions.

$$3+\frac{3}{4}i$$

Real Part

$$3$$

Solution Steps

Multiply both numerator and denominator of \frac{1}{4i} by imaginary unit i.

$$Re(\frac{i}{-4}-2i^{2}-i^{3}+i^{4})$$

Divide i by -4 to get -\frac{1}{4}i.

$$Re(-\frac{1}{4}i-2i^{2}-i^{3}+i^{4})$$

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

$$Re(-\frac{1}{4}i-2\left(-1\right)-i^{3}+i^{4})$$

Multiply 2 and -1 to get -2.

$$Re(-\frac{1}{4}i-\left(-2\right)-i^{3}+i^{4})$$

The opposite of -2 is 2.

$$Re(-\frac{1}{4}i+2-i^{3}+i^{4})$$

Calculate i to the power of 3 and get -i.

$$Re(-\frac{1}{4}i+2-\left(-i\right)+i^{4})$$

The opposite of -i is i.

$$Re(-\frac{1}{4}i+2+i+i^{4})$$

Do the additions in -\frac{1}{4}i+2+i.

$$Re(i^{4}+2+\frac{3}{4}i)$$

Calculate i to the power of 4 and get 1.

$$Re(1+2+\frac{3}{4}i)$$

Do the additions in 1+2+\frac{3}{4}i.

$$Re(3+\frac{3}{4}i)$$

The real part of 3+\frac{3}{4}i is 3.

$$3$$