Solve 5-25i/1+2i-(1-3i)^2

$$\frac { 5 - 25 i } { 1 + 2 i } - ( 1 - 3 i ) ^ { 2 }$$

Evaluate

$$-1-i$$

Solution Steps

Multiply both numerator and denominator of \frac{5-25i}{1+2i} by the complex conjugate of the denominator, 1-2i.

$$\frac{\left(5-25i\right)\left(1-2i\right)}{\left(1+2i\right)\left(1-2i\right)}-\left(1-3i\right)^{2}$$

Do the multiplications in \frac{\left(5-25i\right)\left(1-2i\right)}{\left(1+2i\right)\left(1-2i\right)}.

$$\frac{-45-35i}{5}-\left(1-3i\right)^{2}$$

Divide -45-35i by 5 to get -9-7i.

$$-9-7i-\left(1-3i\right)^{2}$$

Calculate 1-3i to the power of 2 and get -8-6i.

$$-9-7i-\left(-8-6i\right)$$

The opposite of -8-6i is 8+6i.

$$-9-7i+\left(8+6i\right)$$

Add -9-7i and 8+6i to get -1-i.

$$-1-i$$

Real Part

$$-1$$

Solution Steps

Multiply both numerator and denominator of \frac{5-25i}{1+2i} by the complex conjugate of the denominator, 1-2i.

$$Re(\frac{\left(5-25i\right)\left(1-2i\right)}{\left(1+2i\right)\left(1-2i\right)}-\left(1-3i\right)^{2})$$

Do the multiplications in \frac{\left(5-25i\right)\left(1-2i\right)}{\left(1+2i\right)\left(1-2i\right)}.

$$Re(\frac{-45-35i}{5}-\left(1-3i\right)^{2})$$

Divide -45-35i by 5 to get -9-7i.

$$Re(-9-7i-\left(1-3i\right)^{2})$$

Calculate 1-3i to the power of 2 and get -8-6i.

$$Re(-9-7i-\left(-8-6i\right))$$

The opposite of -8-6i is 8+6i.

$$Re(-9-7i+\left(8+6i\right))$$

Add -9-7i and 8+6i to get -1-i.

$$Re(-1-i)$$

The real part of -1-i is -1.

$$-1$$