Compute the following and write in the form x+iy :

[tex]\frac{1+2i}{3-4i} + \frac{2-i}{5i}[/tex]

[tex] \Large{\boxed{\sf \dfrac{1 + 2i}{3 - 4i} + \dfrac{2 - i}{5i } = - \dfrac{2}{5}}} [/tex]

[tex] \\ [/tex]

Given sum:

[tex] \sf \: \dfrac{1+2i}{3-4i} + \dfrac{2-i}{5i}[/tex]

[tex] \\ [/tex]

We can simplify the sum only if the denominators of the two fractions are the same. Since they are different, we have to multiply the numerator and the denominator of each fraction by the denominator of the other one.

[tex] \sf \: \dfrac{1+2i}{3-4i} + \dfrac{2-i}{5i} = \dfrac{5i(1 + 2i)}{5i(3 - 4i)} + \dfrac{(3 - 4i)(2 - i)}{(3 - 4i)5i} \\ \\ \\ \sf \: = \dfrac{5i + 10 {i}^{2} }{15i - 20 {i}^{2} } + \dfrac{6 - 3i - 8i + 4 {i}^{2} }{15i - 20 {i}^{2} } [/tex]

[tex] \\ [/tex]

Replace i² with -1:

[tex] \sf \dfrac{5i + 10 {i}^{2} }{15i - 20 {i}^{2} } + \dfrac{6 - 3i - 8i + 4 {i}^{2} }{15i - 20 {i}^{2} } \: \\ \\ \\ \\ \sf \: = \dfrac{5i + 10( - 1)}{15i - 20( - 1)} + \dfrac{ 6 - 11i + 4( - 1)}{15i - 20( - 1)} \\ \\ \\ \\ \sf = \dfrac{5i - 10}{20 + 15i} + \dfrac{2 - 11i}{20 + 15i} [/tex]

[tex] \\ [/tex]

Simplify the expression:

[tex] \sf = \dfrac{5i - 10}{20 + 15i} + \dfrac{2 - 11i}{20 + 15i} \\ \\ \\ \\ \sf \: = \dfrac{5i - 10 + 2 - 11i}{ 20 + 15i} = \sf \dfrac{ - 8 - 6i}{20 + 15i}[/tex]

[tex] \\ [/tex]

To write our solution in the x + iy form, also known as the algebraic form, we have to understand what the conjugate of a complex number is.

[tex] \textsf{Let z be our complex number, and} \: \overline{\sf z} \: \textsf{its conjugate.} [/tex]

[tex] \\ [/tex]

The conjugate of z, [tex] \overline{ \sf z}, [/tex] is the complex number formed of the same real part as z but of the opposite imaginary part.

Since x is the real part of z, and y is its imaginary part, this can be expressed as:

[tex] \sf If \: z = x + iy \:, then \: \overline{ \sf z} = x - iy [/tex]

[tex] \\ [/tex]

Now, we have to multiple both the denominator and the numerator of our fraction by the conjugate of its denominator:

[tex]\sf \dfrac{ - 8 - 6i}{20 + 15i} = \dfrac{( - 8 - 6i)( \overbrace{20 - 15i}^{ \overline{z}}) }{ (20 + 15i)( \underbrace{20 - 15i}_{ \overline{z}}) } \\ \\ \\ \sf = \dfrac{ - 160 + 120i - 120i + 90 {i}^{2} }{400 - 300i + 300i - 225 {i}^{2} } \\ \\ \\ \sf = \dfrac{ - 160 + 90 {i}^{2} }{400 - 225 {i}^{2} }[/tex]

[tex] \\ [/tex]

One more time, substitute -1 for i²:

[tex] \sf \: \dfrac{ - 160+ 90 {i}^{2} }{400 - 225 {i}^{2} } \: = \dfrac{ - 160 + 90( - 1)}{400 - 225( - 1)} \\ \\ \\ \sf = \boxed{\sf - \dfrac{ 250}{625}} [/tex]

[tex] \\ [/tex]

Finally, let's simplify our result:

[tex] \sf - \dfrac{250}{625} = - \dfrac{2 \times 125}{5 \times 125} = \boxed{ \boxed{ \sf - \dfrac{2}{5}}}[/tex]

[tex] \\ \\ [/tex]

▪️Learn more about the algebraic form of a complex number and the conjugate of a complex number here:

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