Solve [i(2-i)/1+i+i]^2

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

Evaluate

$$\frac{9}{2}i=4.5i$$

Solution Steps

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

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

Multiply both numerator and denominator of \frac{1+2i}{1+i} by the complex conjugate of the denominator, 1-i.

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

Do the multiplications in \frac{\left(1+2i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}.

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

Divide 3+i by 2 to get \frac{3}{2}+\frac{1}{2}i.

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

Add \frac{3}{2}+\frac{1}{2}i and i to get \frac{3}{2}+\frac{3}{2}i.

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

Calculate \frac{3}{2}+\frac{3}{2}i to the power of 2 and get \frac{9}{2}i.

$$\frac{9}{2}i$$

Real Part

$$0$$

Solution Steps

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

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

Multiply both numerator and denominator of \frac{1+2i}{1+i} by the complex conjugate of the denominator, 1-i.

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

Do the multiplications in \frac{\left(1+2i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}.

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

Divide 3+i by 2 to get \frac{3}{2}+\frac{1}{2}i.

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

Add \frac{3}{2}+\frac{1}{2}i and i to get \frac{3}{2}+\frac{3}{2}i.

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

Calculate \frac{3}{2}+\frac{3}{2}i to the power of 2 and get \frac{9}{2}i.

$$Re(\frac{9}{2}i)$$

The real part of \frac{9}{2}i is 0.

$$0$$