Solve sqrt{(1-i/1+i+2i)^2}

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

Evaluate

$$i$$

Solution Steps

Multiply both numerator and denominator of \frac{1-i}{1+i} by the complex conjugate of the denominator, 1-i.

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

Do the multiplications in \frac{\left(1-i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}.

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

Divide -2i by 2 to get -i.

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

Add -i and 2i to get i.

$$\sqrt{i^{2}}$$

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

$$\sqrt{-1}$$

Calculate the square root of -1 and get i.

$$i$$

Real Part

$$0$$

Solution Steps

Multiply both numerator and denominator of \frac{1-i}{1+i} by the complex conjugate of the denominator, 1-i.

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

Do the multiplications in \frac{\left(1-i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}.

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

Divide -2i by 2 to get -i.

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

Add -i and 2i to get i.

$$Re(\sqrt{i^{2}})$$

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

$$Re(\sqrt{-1})$$

Calculate the square root of -1 and get i.

$$Re(i)$$

The real part of i is 0.

$$0$$