Complex number 4

Algebra Level 3

Simplify:

3 + 2 i 2 + 3 i + 1 + 5 i 1 − 2 i \large \frac{3+2i}{2+3i} + \frac{1+5i}{1-2i}

Notation: i = − 1 i=\sqrt{-1} denotes the imaginary unit .

see also: Complex number

− 72 65 + 47 65 i \dfrac{-72}{65} + \dfrac{47}{65}i 6 i 6i − 57 65 + 66 65 i \dfrac{-57}{65} + \dfrac{66}{65}i − 57 65 + 64 65 i \dfrac{-57}{65} + \dfrac{64}{65}i 58 65 + 66 65 i \dfrac{58}{65} + \dfrac{66}{65}i

Nazmus Sakib by Nazmus sakib , 24, Bangladesh

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

3 + 2 i 2 + 3 i + 1 + 5 i 1 − 2 i = ( 3 + 2 i ) ( 2 − 3 i ) ( 2 + 3 i ) ( 2 − 3 i ) + ( 1 + 5 i ) ( 1 + 2 i ) ( 1 − 2 i ) ( 1 + 2 i ) = 12 − 5 i 13 + − 9 + 7 i 5 = 60 − 25 i 65 + − 117 + 91 i 65 = − 57 65 + 66 i 65 \begin{aligned} \frac {3+2i}{2+3i} + \frac {1+5i}{1-2i} & = \frac {(3+2i)(2-3i)}{(2+3i)(2-3i)} + \frac {(1+5i)(1+2i)}{(1-2i)(1+2i)} \\ & = \frac {12-5i}{13} + \frac {-9+7i}5 \\ & = \frac {60-25i}{65} + \frac {-117+91i}{65} \\ & = \boxed{\dfrac {-57}{65} + \dfrac {66i}{65}} \end{aligned}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Is it mandatory to rationalise the numerator and denominator?

Kaushik Chandra - 3 years, 4 months ago

Log in to reply

No, I don't think so.

Chew-Seong Cheong - 3 years, 4 months ago

That's pretty easy

Shruti Yadav - 3 years, 7 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...