Punch
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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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Punch said:Given \(z=r(cos\theta+isin\theta)\), solve for \(z\) in the form \(a+ib\) if \(4+i(4z+1)=2re^{i(\pi+\theta)}\)
Punch said:Given \(z=r(cos\theta+isin\theta)\), solve for \(z\) in the form \(a+ib\) if \(4+i(4z+1)=2re^{i(\pi+\theta)}\)
CaptainBlack said: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 said:Given \(z=r(cos\theta+isin\theta)\), solve for \(z\) in the form \(a+ib\) if \(4+i(4z+1)=2re^{i(\pi+\theta)}\)
Prove It said:\[ \displaystyle \begin{align*} 4 + i\left(4z + 1\right) &= 2r\,e^{i\left(\pi + \theta\right)} \\ 4 + i\left(4z + 1\right) &= 2r\left[\cos{\left( \pi + \theta \right)} + i\sin{\left( \pi + \theta \right)} \right] \\ 4 + i\left(4z + 1\right) &= 2r\left[-\cos{\left(\theta\right)} - i\sin{\left(\theta\right)}\right] \\ 4 + i\left(4z + 1\right) &= -2r\left[\cos{\left(\theta\right)} + i\sin{\left(\theta\right)} \right] \\ 4 + i\left(4z + 1\right) &= -2z \end{align*} \]
Now solve for $\displaystyle z$.
Punch said:I see that there is another method to solve. But could u also show me the method of solving for theta and r?
Punch said:I see that there is another method to solve. But could u also show me the method of solving for theta and r?
CaptainBlack said: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 said:Really? Because that was my first instinct as well, and ended up with two equations in four unknowns...
CaptainBlack said: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
CaptainBlack said: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
Punch said:First, I am 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 couldn't 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.
CaptainBlack said: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
CaptainBlack said: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 said: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?