Solve (1+i)^2*1/(1-2i)=

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

Evaluate

$$-\frac{4}{5}+\frac{2}{5}i=-0.8+0.4i$$

Solution Steps

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

$$2i\times \frac{1}{1-2i}$$

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

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

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

$$2i\times \frac{1+2i}{5}$$

Divide 1+2i by 5 to get \frac{1}{5}+\frac{2}{5}i.

$$2i\left(\frac{1}{5}+\frac{2}{5}i\right)$$

Multiply 2i and \frac{1}{5}+\frac{2}{5}i to get -\frac{4}{5}+\frac{2}{5}i.

$$-\frac{4}{5}+\frac{2}{5}i$$

Real Part

$$-\frac{4}{5} = -0.8$$

Solution Steps

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

$$Re(2i\times \frac{1}{1-2i})$$

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

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

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

$$Re(2i\times \frac{1+2i}{5})$$

Divide 1+2i by 5 to get \frac{1}{5}+\frac{2}{5}i.

$$Re(2i\left(\frac{1}{5}+\frac{2}{5}i\right))$$

Multiply 2i and \frac{1}{5}+\frac{2}{5}i to get -\frac{4}{5}+\frac{2}{5}i.

$$Re(-\frac{4}{5}+\frac{2}{5}i)$$

The real part of -\frac{4}{5}+\frac{2}{5}i is -\frac{4}{5}.

$$-\frac{4}{5}$$