Division with the rectangular form

AI Thread Summary
The discussion revolves around solving the division of complex numbers without using polar transformation. The example provided is (2 + 2i) / (1 - i). The correct method involves multiplying both the numerator and denominator by the complex conjugate of the denominator, resulting in (2 + 2i)(1 + i) in the numerator and (1 - i)(1 + i) in the denominator. This leads to a final answer of 2i, confirming the calculation is correct. The participant expresses gratitude for the clarification after initially feeling confused.
Truthlover
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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: \frac{2+2i}{1-i}

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

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
 
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Multiply both numerator and denominator with the complex conjugate of (1 - i ).
 
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?
 
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Truthlover said:
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!

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

\frac{2+2i}{1- i}= \frac{4i}{2}= 2i

which is correct:
(2i)(1- i)= 2i- 2i^2= 2+ 2i.
 
I feel really stupid but thanks a lot
 
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