# complex numbers II

#### Punch

##### New member
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
Re: complex numbers

Given $$z=r(cos\theta+isin\theta)$$, solve for $$z$$ in the form $$a+ib$$ if $$4+i(4z+1)=2re^{i(\pi+\theta)}$$

#### Punch

##### New member
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
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
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

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

• Punch