Dividing Complex Numbers

[nycmathguy]

Section 6.6

 

[MathLover1]

you have

, that is incorrect

need to do this way:

=

=

...........rationalize

=

=

=

=0.086824+0.492405*i

you did same mistake with

do it like I did above and result should be:
-0.735626 +1.01062*i

 

[nycmathguy]

you have

, that is incorrect

need to do this way:

=

=

...........rationalize

=

=

=

=0.086824+0.492405*i

you did same mistake with

do it like I did above and result should be:
-0.735626 +1.01062*i

Did I get both questions wrong? What error did I make twice?

 

[MathLover1]

yes

 

[nycmathguy]

yes

What error did I make twice?

 

[MathLover1]

you cannot factor out 1/2 and write cos and sin as difference of angles from numerator and denominator
that is WRONG

compare to (cos(x)+sin(x))/(cos(y)+sin(y))-> you cannot write it as cos(x-y)+sin(x-y)

 

[nycmathguy]

you cannot factor out 1/2 and write cos and sin as difference of angles from numerator and denominator
that is WRONG

compare to (cos(x)+sin(x))/(cos(y)+sin(y))-> you cannot write it as cos(x-y)+sin(x-y)

Let me post what Ron Larson said to do. BRB.

 

[nycmathguy]

you cannot factor out 1/2 and write cos and sin as difference of angles from numerator and denominator
that is WRONG

compare to (cos(x)+sin(x))/(cos(y)+sin(y))-> you cannot write it as cos(x-y)+sin(x-y)

What do you say about this explanation by Ron Larson? I see subtraction by Larson. No?

 

[MathLover1]

that is new to me, I do not recall that method but it's much shorter (I was always doing rationalization)

I know the theorem for division of complex numbers is:

Let \( z1=(r1,θ1)\) and \(z2=(r2,θ2)\) be complex numbers expressed in polar form, such that z2≠0.

Then:
\(z1/z2=r1/r2(cos(θ1-θ2)+i*sin(θ1-θ2))\)

Proof
\(z1/z2 =r1(cosθ1+isinθ1/r2(cosθ2+isinθ2)\).Definition of Polar Form of Complex Number

=\((r1(cosθ1+isinθ1))(r2(cosθ2−isinθ2))(r2(cosθ2+isinθ2)(r2(cosθ2−isinθ2))\).multiplying numerator and denominator by \(r2(cosθ1-isinθ1) \)

=\(r1r2(cos(θ1−θ2)+isin(θ1−θ2))/(r2^2(cos(θ2−θ2)+isin(θ2θ2))\).Product of Complex Numbers in Polar Form

=\( (r1(cos(θ1−θ2)+isin(θ1−θ2)))/(r2(cos0+isin0))\)

=\((r1/r2)(cos(θ1−θ2+isin(θ1-θ2)) \) ........Cosine of Zero is One and Sine of Zero is Zero

 

[nycmathguy]

that is new to me, I do not recall that method but it's much shorter (I was always doing rationalization)

I know the theorem for division of complex numbers is:

Let z1=(r1,θ1) and z2=(r2,θ2) be complex numbers expressed in polar form, such that z2≠0.

Then:
\(z1/z2=r1/r2(cos(θ1-θ2)+i*sin(θ1-θ2))\)

Proof
z1/z2 =r1(cosθ1+isinθ1/r2(cosθ2+isinθ2)..........Definition of Polar Form of Complex Number

=(r1(cosθ1+isinθ1))(r2(cosθ2−isinθ2))(r2(cosθ2+isinθ2))(r2(cosθ2−isinθ2))................multiplying numerator and denominator by r2(cosθ1-isinθ1)

=r1r2(cos(θ1−θ2)+isin(θ1−θ2))/(r2^2(cos(θ2−θ2)+isin(θ2θ2))....Product of Complex Numbers in Polar Form

= (r1(cos(θ1−θ2)+isin(θ1−θ2)))/(r2(cos0+isin0))

=(r1/r2)(cos(θ1−θ2+isin(θ1-θ2)) ........Cosine of Zero is One and Sine of Zero is Zero

Is Larson wrong?

 

[MathLover1]

this is a result using wolfram alpha

 

[nycmathguy]

this is a result using wolfram alpha

What do I do? Unfortunately, the answers to even number problems are not listed in the back of the book as you already know. In that case, why not do 41 and 43? I will then check the answer listed on the back of the book by Larson. Sounds good?

 

[MathLover1]

ok

 

[Country Boy]

To divide $\frac{a+ bi}{c+ di}$, multiply both numerator and denominator by the conjugate of the denominator:

 

[nycmathguy]

To divide $\frac{a+ bi}{c+ di}$, multiply both numerator and denominator by the conjugate of the denominator:

Ok. I will try this as well.