complex number fraction

[Quasimoto]

Dear forum users,

I'm not sure if I am at the right place to ask this question but here it goes.
Complex numbers are new to me and I'm just starting to learn how to work with them.
I have to work out this problem (fig 1.1) and write it in the the form of x+yi.

(fig. 1.1)


I do know that if you have a problem that looks like (fig. 1.2) you have to multiply with the conjugate to remove the i the denominator.
But with this problem I can't figure out where to start.

(fig. 1.2)


I hope someone can help me out or at least get me started, your help is appreciated.

 

[Dr.Peterson]

Multiply the numerator and denominator by BOTH conjugates, namely (2-i)(1+3i). The rest will just be a long slog through several products, multiplying one complex number by another at a time. The denominator will be relatively easy, of course.

If you get stuck, show us how far you get and we may have some hints for making the work more bearable.

 

[Romsek]

Just take it in two steps

[MATH] \dfrac{(2+3i)(-1+i)}{(2+i)(1-3i)} = \\ \dfrac{(2+3i)(-1+i)}{(2+i)(1-3i)} \cdot \dfrac{2-i}{2-i} = \\ \dfrac{(2+3i)(-1+i)(2-i)}{5(1-3i)} = \\ \dfrac{(2+3i)(-1+i)(2-i)}{5(1-3i)} \cdot \dfrac{1+3i}{1+3i} = \\ \dfrac{(2+3i)(-1+i)(2-i)(1+3i)}{5\cdot 10} = \\ \dfrac{(2+3i)(-1+i)(2-i)(1+3i)}{50} [/MATH]
and I leave you to expand all that out and collect it back into \(\displaystyle x+ i y\) form.

 

[Deleted member 4993]

Another similar way:

Multiply the numerator and denominator out:

\(\displaystyle \frac{(2+3i)(-1+i)}{(2+i)(1-3i)}\)

=\(\displaystyle \frac{2*(-1+i)+3i*(-1+i)}{2*(1-3i)+i*(1-3i)}\)

=\(\displaystyle \frac{-2 + 2i - 3i - 3}{2 - 6i + i + 3}\)

=\(\displaystyle \frac{-5 - i}{5 - 5i}\)........................Now multiply & divide by the conjugate of the denominator

=\(\displaystyle \frac{(-5 - i)*(5 + 5i)}{(5 - 5i) * (5 + 5i)}\) ..................... and finish it..............

 

[MarkFL]

View attachment 16819

Another similar way:

Multiply the numerator and denominator out:

\(\displaystyle \frac{(2+3i)(-1+i)}{(2+i)(1-3i)}\)

=\(\displaystyle \frac{2*(-1+i)+3i*(-1+i)}{2*(1-3i)+i*(1-3i)}\)

=\(\displaystyle \frac{-2 + 2i - 3i - 3}{2 - 6i + i + 3}\)

=\(\displaystyle \frac{-5 - i}{5 - 5i}\)........................Now multiply & divide by the conjugate of the denominator

=\(\displaystyle \frac{(-5 - i)*(5 + 5i)}{(5 - 5i) * (5 + 5i)}\) ..................... and finish it..............

That's how I'd be inclined to approach it.