Solve (2+i)*z-(3-2i)/2z=4+3i-(2-5i)z

$$( 2 + i ) \cdot z - \frac { ( 3 - 2 i ) } { 2 } z = 4 + 3 i - ( 2 - 5 i ) z$$

Solve for z

$$z=\frac{4}{61}+\frac{78}{61}i\approx 0.06557377+1.278688525i$$

Steps for Solving Linear Equation

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

$$\left(2+i\right)z-\left(\frac{3}{2}-i\right)z=4+3i-\left(2-5i\right)z$$

Combine \left(2+i\right)z and \left(-\frac{3}{2}+i\right)z to get \left(\frac{1}{2}+2i\right)z.

$$\left(\frac{1}{2}+2i\right)z=4+3i-\left(2-5i\right)z$$

Add \left(2-5i\right)z to both sides.

$$\left(\frac{1}{2}+2i\right)z+\left(2-5i\right)z=4+3i$$

Combine \left(\frac{1}{2}+2i\right)z and \left(2-5i\right)z to get \left(\frac{5}{2}-3i\right)z.

$$\left(\frac{5}{2}-3i\right)z=4+3i$$

Divide both sides by \frac{5}{2}-3i.

$$z=\frac{4+3i}{\frac{5}{2}-3i}$$

Multiply both numerator and denominator of \frac{4+3i}{\frac{5}{2}-3i} by the complex conjugate of the denominator, \frac{5}{2}+3i.

$$z=\frac{\left(4+3i\right)\left(\frac{5}{2}+3i\right)}{\left(\frac{5}{2}-3i\right)\left(\frac{5}{2}+3i\right)}$$

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

$$z=\frac{\left(4+3i\right)\left(\frac{5}{2}+3i\right)}{\left(\frac{5}{2}\right)^{2}-3^{2}i^{2}}$$

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

$$z=\frac{\left(4+3i\right)\left(\frac{5}{2}+3i\right)}{\frac{61}{4}}$$

Multiply complex numbers 4+3i and \frac{5}{2}+3i like you multiply binomials.

$$z=\frac{4\times \frac{5}{2}+4\times \left(3i\right)+3i\times \frac{5}{2}+3\times 3i^{2}}{\frac{61}{4}}$$

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

$$z=\frac{4\times \frac{5}{2}+4\times \left(3i\right)+3i\times \frac{5}{2}+3\times 3\left(-1\right)}{\frac{61}{4}}$$

Do the multiplications in 4\times \frac{5}{2}+4\times \left(3i\right)+3i\times \frac{5}{2}+3\times 3\left(-1\right).

$$z=\frac{10+12i+\frac{15}{2}i-9}{\frac{61}{4}}$$

Combine the real and imaginary parts in 10+12i+\frac{15}{2}i-9.

$$z=\frac{10-9+\left(12+\frac{15}{2}\right)i}{\frac{61}{4}}$$

Do the additions in 10-9+\left(12+\frac{15}{2}\right)i.

$$z=\frac{1+\frac{39}{2}i}{\frac{61}{4}}$$

Divide 1+\frac{39}{2}i by \frac{61}{4} to get \frac{4}{61}+\frac{78}{61}i.

$$z=\frac{4}{61}+\frac{78}{61}i$$