What is (3+6i)+(4-i)(3+5i)+1/(2-i) ?

The solution to (3+6i)+(4-i)(3+5i)+1/(2-i) is 102/5+116/5 i

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simplify (3+6i)+(4-i)(3+5i)+1/(2-i)

simplify

(3 + 6 i) + (4 - i) (3 + 5 i) + \frac{1}{2 - i}

\frac{102}{5} + \frac{116}{5} i

Solution steps

(3 + 6 i) + (4 - i) (3 + 5 i) + \frac{1}{2 - i}

Apply rule:

(a) = a

(3 + 6 i) = 3 + 6 i

= 3 + 6 i + (4 - i) (3 + 5 i) + \frac{1}{2 - i}

(4 - i) (3 + 5 i) = 17 + 17 i

\frac{1}{2 - i} = \frac{2 + i}{5}

= 3 + 6 i + 17 + 17 i + \frac{2 + i}{5}

Group like terms

= 6 i + 17 i + \frac{2 + i}{5} + 3 + 17

Add similar elements:

6 i + 17 i = 23 i

= 23 i + \frac{2 + i}{5} + 3 + 17

Add the numbers:

3 + 17 = 20

= 23 i + \frac{2 + i}{5} + 20

Join

23 i + \frac{2 + i}{5} + 20 : \frac{116 i + 102}{5}

= \frac{116 i + 102}{5}

Rewrite in standard complex form:

\frac{102}{5} + \frac{116}{5} i

= \frac{102}{5} + \frac{116}{5} i

\frac{2 π}{\frac{π}{4}}

980 \cdot \frac{3}{7}

10 x - 5 x^{2} = \frac{10}{3}

- x^{2} + 14 x - 49

x^{2} = 9 x - 20

What is (3+6i)+(4-i)(3+5i)+1/(2-i) ?
The solution to (3+6i)+(4-i)(3+5i)+1/(2-i) is 102/5+116/5 i