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complex numbers II

Punch

New member
Jan 29, 2012
23
Given \(z=r(cos\theta+isin\theta)\), solve for \(z\) in the form \(a+ib\) if \(4+i(4z+1)=2re^{i(\pi+\theta)}\)
 
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CaptainBlack

Well-known member
Jan 26, 2012
890
Re: complex numbers

Given \(z=r(cos\theta+isin\theta)\), solve for \(z\) in the form \(a+ib\) if \(4+i(4z+1)=2re^{i(\pi+\theta)}\)
Use \$ delimiters around in-line LaTeX maths and \$\$ delimiters around display LaTeX maths.

CB
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Re: complex numbers

Given \(z=r(cos\theta+isin\theta)\), solve for \(z\) in the form \(a+ib\) if \(4+i(4z+1)=2re^{i(\pi+\theta)}\)
Expand both sides in Cartesian form, remembering that \(e^{i\phi}=\cos(\phi)+i \sin(\phi)\), then equate real and imaginary parts and solve for \(r\) and \( \theta \)

Alternative method: rewrite the left hand side in polar form and equate modulus and arguments.

CB
 

Punch

New member
Jan 29, 2012
23
Re: complex numbers

Expand both sides in Cartesian form, remembering that \(e^{i\phi}=\cos(\phi)+i \sin(\phi)\), then equate real and imaginary parts and solve for \(r\) and \( \theta \)

Alternative method: rewrite the left hand side in polar form and equate modulus and arguments.

CB
One question, do i expand z too?

For the cartesian approach yes, it then will give you a pair of simultaneous equations in \(r\) and \(\theta\) to solve.

CB
 
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Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Given \(z=r(cos\theta+isin\theta)\), solve for \(z\) in the form \(a+ib\) if \(4+i(4z+1)=2re^{i(\pi+\theta)}\)
\[ \displaystyle \begin{align*} 4 + i\left(4z + 1\right) &= 2r\,e^{i\left(\pi + \theta\right)} \\ 4 + i\left(4z + 1\right) &= 2r\left[\cos{\left( \pi + \theta \right)} + i\sin{\left( \pi + \theta \right)} \right] \\ 4 + i\left(4z + 1\right) &= 2r\left[-\cos{\left(\theta\right)} - i\sin{\left(\theta\right)}\right] \\ 4 + i\left(4z + 1\right) &= -2r\left[\cos{\left(\theta\right)} + i\sin{\left(\theta\right)} \right] \\ 4 + i\left(4z + 1\right) &= -2z \end{align*} \]

Now solve for $\displaystyle z$.
 

Punch

New member
Jan 29, 2012
23
\[ \displaystyle \begin{align*} 4 + i\left(4z + 1\right) &= 2r\,e^{i\left(\pi + \theta\right)} \\ 4 + i\left(4z + 1\right) &= 2r\left[\cos{\left( \pi + \theta \right)} + i\sin{\left( \pi + \theta \right)} \right] \\ 4 + i\left(4z + 1\right) &= 2r\left[-\cos{\left(\theta\right)} - i\sin{\left(\theta\right)}\right] \\ 4 + i\left(4z + 1\right) &= -2r\left[\cos{\left(\theta\right)} + i\sin{\left(\theta\right)} \right] \\ 4 + i\left(4z + 1\right) &= -2z \end{align*} \]

Now solve for $\displaystyle z$.
I see that there is another method to solve. But could u also show me the method of solving for theta and r?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
I see that there is another method to solve. But could u also show me the method of solving for theta and r?
No.
1) It's longer.
2) It's much more difficult.
3) It's pointless when there is a much more straightforward method such as what I have shown you.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
I see that there is another method to solve. But could u also show me the method of solving for theta and r?
Just write the equations out in Cartesian form, the solution from that point is easier than you might think once you see the form of the equations.

CB
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Just write the equations out in Cartesian form, the solution from that point is easier than you might think once you see the form of the equations.

CB
Really? Because that was my first instinct as well, and ended up with two equations in four unknowns...
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Really? Because that was my first instinct as well, and ended up with two equations in four unknowns...
There are only two unknowns \(r\) and \(\theta\) (even in the Cartesian form of the equations). Eliminate \(r\) by dividing one by the other (after a bit of rearrangement) and you are left with a linear equation in \(\tan(\theta)\) (at least after dividing top and bottom by \(\cos(\theta)\) )

On second thoughts we are talking at cross purposes, I think what you mean is to write the equation as:

\(4+i(4z+1)=-2z\)

and then solve for \(z\) by common algebra and then simplify.

CB
 
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Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
There are only two unknowns \(r\) and \(\theta\) (even in the Cartesian form of the equations). Eliminate \(r\) by dividing one by the other (after a bit of rearrangement) and you are left with a linear equation in \(\tan(\theta)\) (at least after dividing top and bottom by \(\cos(\theta)\) )

On second thoughts we are talking at cross purposes, I think what you mean is to write the equation as:

\(4+i(4z+1)=-2z\)

and then solve for \(z\) by common algebra and then simplify.

CB
I know what I did wrong now, I had forgotten that we were given $\displaystyle z = r\left[\cos{\left(\theta\right)} + i\sin{\left(\theta\right)}\right]$.
 

Punch

New member
Jan 29, 2012
23
There are only two unknowns \(r\) and \(\theta\) (even in the Cartesian form of the equations). Eliminate \(r\) by dividing one by the other (after a bit of rearrangement) and you are left with a linear equation in \(\tan(\theta)\) (at least after dividing top and bottom by \(\cos(\theta)\) )

On second thoughts we are talking at cross purposes, I think what you mean is to write the equation as:

\(4+i(4z+1)=-2z\)

and then solve for \(z\) by common algebra and then simplify.

CB
First, im sorry as i had a problem using the latex... Please do me a favour by helping me to add the latex

Secondly, I tried using this approach and formed 2 equations \( r=\frac{sin(\pi+\theta)-1}{4cos\theta}\) and \(r=\frac{2}{cos9\pi+\theta0+2sin\theta}\)

Dividing one by the other, I formed the equation \(\frac{8}{(-1+2tan\theta)(-sin\theta-1)}=1\) which i couldnt solve for \(\theta\)...

The method is also very long. Did i do it the right way? Please show me the right way if i did not.
 
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CaptainBlack

Well-known member
Jan 26, 2012
890
First, im sorry as i had a problem using the latex... Please do me a favour by helping me to add the latex

Secondly, I tried using this approach and formed 2 equations \( r=\frac{sin(\pi+\theta)-1}{4cos\theta}\) and \(r=\frac{2}{cos9\pi+\theta0+2sin\theta}\)

Dividing one by the other, I formed the equation \(\frac{8}{(-1+2tan\theta)(-sin\theta-1)}=1\) which i couldnt solve for \(\theta\)...

The method is also very long. Did i do it the right way? Please show me the right way if i did not.
The equations I get are:

\[4r \cos(\theta)+2r \sin(\theta)=-1\] \[2r \cos(\theta)-4r\sin(\theta)=-4\]

Which when divided give:

\[\frac{2 \cos(\theta)+\sin(\theta)}{\cos(\theta)-2\sin(\theta)}=\frac{1}{4}\]

CB
 

Punch

New member
Jan 29, 2012
23
The equations I get are:

\[4r \cos(\theta)+2r \sin(\theta)=-1\] \[2r \cos(\theta)-4r\sin(\theta)=-4\]

Which when divided give:

\[\frac{2 \cos(\theta)+\sin(\theta)}{\cos(\theta)-2\sin(\theta)}=\frac{1}{4}\]

CB
I will try it again, thanks

---------- Post added at 06:49 PM ---------- Previous post was at 06:19 PM ----------

The equations I get are:

\[4r \cos(\theta)+2r \sin(\theta)=-1\] \[2r \cos(\theta)-4r\sin(\theta)=-4\]

Which when divided give:

\[\frac{2 \cos(\theta)+\sin(\theta)}{\cos(\theta)-2\sin(\theta)}=\frac{1}{4}\]

CB
You're right. However, solving for theta is not the end of the story... The question requires the answer to be presented in the form a+ib. Am I right in saying that tackling the question using this method would take very long? Or is there another way to find a and b after solving for r and theta?
 

CaptainBlack

Well-known member
Jan 26, 2012
890



You're right. However, solving for theta is not the end of the story... The question requires the answer to be presented in the form a+ib. Am I right in saying that tackling the question using this method would take very long? Or is there another way to find a and b after solving for r and theta?
\(a=r \cos(\theta),\ b=r \sin(\theta)\)

CB