Solve 91quad((3+2i)/(-2-5i))

$$91 \quad ( \frac { ( 3 + 2 i ) } { ( - 2 - 5 i ) } )$$

Evaluate

$$-\frac{1456}{29}+\frac{1001}{29}i\approx -50.206896552+34.517241379i$$

Solution Steps

Multiply both numerator and denominator of \frac{3+2i}{-2-5i} by the complex conjugate of the denominator, -2+5i.

$$91\times \frac{\left(3+2i\right)\left(-2+5i\right)}{\left(-2-5i\right)\left(-2+5i\right)}$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$91\times \frac{\left(3+2i\right)\left(-2+5i\right)}{\left(-2\right)^{2}-5^{2}i^{2}}$$

By definition, i^{2} is -1. Calculate the denominator.

$$91\times \frac{\left(3+2i\right)\left(-2+5i\right)}{29}$$

Multiply complex numbers 3+2i and -2+5i like you multiply binomials.

$$91\times \frac{3\left(-2\right)+3\times \left(5i\right)+2i\left(-2\right)+2\times 5i^{2}}{29}$$

By definition, i^{2} is -1.

$$91\times \frac{3\left(-2\right)+3\times \left(5i\right)+2i\left(-2\right)+2\times 5\left(-1\right)}{29}$$

Do the multiplications in 3\left(-2\right)+3\times \left(5i\right)+2i\left(-2\right)+2\times 5\left(-1\right).

$$91\times \frac{-6+15i-4i-10}{29}$$

Combine the real and imaginary parts in -6+15i-4i-10.

$$91\times \frac{-6-10+\left(15-4\right)i}{29}$$

Do the additions in -6-10+\left(15-4\right)i.

$$91\times \frac{-16+11i}{29}$$

Divide -16+11i by 29 to get -\frac{16}{29}+\frac{11}{29}i.

$$91\left(-\frac{16}{29}+\frac{11}{29}i\right)$$

Multiply 91 times -\frac{16}{29}+\frac{11}{29}i.

$$91\left(-\frac{16}{29}\right)+91\times \left(\frac{11}{29}i\right)$$

Do the multiplications.

$$-\frac{1456}{29}+\frac{1001}{29}i$$

Real Part

$$-\frac{1456}{29} = -50\frac{6}{29} = -50.206896551724135$$

Solution Steps

Multiply both numerator and denominator of \frac{3+2i}{-2-5i} by the complex conjugate of the denominator, -2+5i.

$$Re(91\times \frac{\left(3+2i\right)\left(-2+5i\right)}{\left(-2-5i\right)\left(-2+5i\right)})$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$Re(91\times \frac{\left(3+2i\right)\left(-2+5i\right)}{\left(-2\right)^{2}-5^{2}i^{2}})$$

By definition, i^{2} is -1. Calculate the denominator.

$$Re(91\times \frac{\left(3+2i\right)\left(-2+5i\right)}{29})$$

Multiply complex numbers 3+2i and -2+5i like you multiply binomials.

$$Re(91\times \frac{3\left(-2\right)+3\times \left(5i\right)+2i\left(-2\right)+2\times 5i^{2}}{29})$$

By definition, i^{2} is -1.

$$Re(91\times \frac{3\left(-2\right)+3\times \left(5i\right)+2i\left(-2\right)+2\times 5\left(-1\right)}{29})$$

Do the multiplications in 3\left(-2\right)+3\times \left(5i\right)+2i\left(-2\right)+2\times 5\left(-1\right).

$$Re(91\times \frac{-6+15i-4i-10}{29})$$

Combine the real and imaginary parts in -6+15i-4i-10.

$$Re(91\times \frac{-6-10+\left(15-4\right)i}{29})$$

Do the additions in -6-10+\left(15-4\right)i.

$$Re(91\times \frac{-16+11i}{29})$$

Divide -16+11i by 29 to get -\frac{16}{29}+\frac{11}{29}i.

$$Re(91\left(-\frac{16}{29}+\frac{11}{29}i\right))$$

Multiply 91 times -\frac{16}{29}+\frac{11}{29}i.

$$Re(91\left(-\frac{16}{29}\right)+91\times \left(\frac{11}{29}i\right))$$

Do the multiplications in 91\left(-\frac{16}{29}\right)+91\times \left(\frac{11}{29}i\right).

$$Re(-\frac{1456}{29}+\frac{1001}{29}i)$$

The real part of -\frac{1456}{29}+\frac{1001}{29}i is -\frac{1456}{29}.

$$-\frac{1456}{29}$$