Solve 1+9i/(2+i)^2=

$$\frac { 1 + 9 i } { ( 2 + i ) ^ { 2 } } =$$

Evaluate

$$\frac{39}{25}+\frac{23}{25}i=1.56+0.92i$$

Solution Steps

Calculate 2+i to the power of 2 and get 3+4i.

$$\frac{1+9i}{3+4i}$$

Multiply both numerator and denominator by the complex conjugate of the denominator, 3-4i.

$$\frac{\left(1+9i\right)\left(3-4i\right)}{\left(3+4i\right)\left(3-4i\right)}$$

Do the multiplications in \frac{\left(1+9i\right)\left(3-4i\right)}{\left(3+4i\right)\left(3-4i\right)}.

$$\frac{39+23i}{25}$$

Divide 39+23i by 25 to get \frac{39}{25}+\frac{23}{25}i.

$$\frac{39}{25}+\frac{23}{25}i$$

Real Part

$$\frac{39}{25} = 1\frac{14}{25} = 1.56$$

Solution Steps

Calculate 2+i to the power of 2 and get 3+4i.

$$Re(\frac{1+9i}{3+4i})$$

Multiply both numerator and denominator of \frac{1+9i}{3+4i} by the complex conjugate of the denominator, 3-4i.

$$Re(\frac{\left(1+9i\right)\left(3-4i\right)}{\left(3+4i\right)\left(3-4i\right)})$$

Do the multiplications in \frac{\left(1+9i\right)\left(3-4i\right)}{\left(3+4i\right)\left(3-4i\right)}.

$$Re(\frac{39+23i}{25})$$

Divide 39+23i by 25 to get \frac{39}{25}+\frac{23}{25}i.

$$Re(\frac{39}{25}+\frac{23}{25}i)$$

The real part of \frac{39}{25}+\frac{23}{25}i is \frac{39}{25}.

$$\frac{39}{25}$$