How can you simplify complex division with imaginary numbers?

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To simplify the complex division (1+2i+3i²)/(1-2i+3i²), first replace i² with -1 in both the numerator and denominator. This transforms the expression into a standard form of a complex number. Next, multiply both the numerator and denominator by the conjugate of the denominator to eliminate the imaginary part in the denominator. After simplification, the result is -i, which matches the correct answer provided in the book. This method effectively clarifies the process of handling complex numbers in division.
alijan kk
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Homework Statement


(1+2i+3i2)/(1-2i+3i2)

answer options : a : 1 b: -i c: i d: 0

Homework Equations


what is the most easy method to solve it ,

The Attempt at a Solution


are they conjugate to each other ? if they are than z/zconjugate =1 ,
but how can i make shure that they are conjugate to each other
[/B]
 
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alijan kk said:

Homework Statement


(1+2i+3i2)/(1-2i+3i2)

answer options : a : 1 b: -i c: i d: 0

Homework Equations


what is the most easy method to solve it ,
Simplify the ##i^2## term in the numerator and denominator, and then multiply both numerator and denominator by the conjugate of the denominator. You should already know that ##i^2 = -1##.
alijan kk said:

The Attempt at a Solution


are they conjugate to each other ? if they are than z/zconjugate =1 , [/B]
That's not true. ##\frac z {\bar z} \neq 1## unless z is purely real.
alijan kk said:
but how can i make shure that they are conjugate to each other
 
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Step 1, convert each of the numerator and denominator into the form ##a+bi## by replacing ##i^2## by a number that doesn't involve ##i## in both, then collecting terms and simplifying.
Step 2: Make the denominator real by multiplying both the numerator and the denominator by the conjugate of the denominator.
Step 3: simplify.
 
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Mark44 said:
Simplify the ##i^2## term in the numerator and denominator, and then multiply both numerator and denominator by the conjugate of the denominator. You should already know that ##i^2 = -1##.
That's not true. ##\frac z {\bar z} \neq 1## unless z is purely real.
i simplified the equation and i got (1-i)/(1+i) and by dividing it I got -i. which is the correct answer in the book , thankyou.
 

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