Division with the rectangular form

[Truthlover]

Hi everyone, I was questionning myself about a problem that I have surely learn in school but I want to know if it's possible to solve a division with imaginary numbers without using the polar transformation.

Example: [tex]\frac{2+2i}{1-i}[/tex]

So with the polar tansformation we have this:[tex]\frac{2\sqrt{2}\angle45°}{\sqrt{2}\angle-45°}[/tex][tex]=2i[/tex]

Now I was wondering if someone know a way to find the solution of 2i without the polar transformation. If it's the case can you show me how you have done it.


Thanks

 

[Svalbard]

Multiply both numerator and denominator with the complex conjugate of (1 - i ).

 

[Truthlover]

I'm not sure if I do the right thing but the conugate of (1-i) is (1+i). So if we do the multiplaction it give: (2+2i)*(1+i)= (2*1)+(2*i)+(2i*1)+(2i*i)= 2+2i+2i-2=4i

This is not the answer. What I have done wrong?

 

[HallsofIvy]

I'm not sure if I do the right thing but the conugate of (1-i) is (1+i). So if we do the multiplaction it give: (2+2i)*(1+i)= (2*1)+(2*i)+(2i*1)+(2i*i)= 2+2i+2i-2=4i

This is not the answer. What I have done wrong?

You didn't include the denominator!

[tex]\frac{2+2i}{1-i}\frac{1+i}{1+i}= \frac{(2+2i)(1+i)}{(1-i)(1+i)}[/tex]
Now the numerator is, as you say, 4i. The denominator is 1- i²= 2.

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

which is correct:
[tex](2i)(1- i)= 2i- 2i^2= 2+ 2i[/tex].

 

[Truthlover]

I feel really stupid but thanks a lot