Solution of a Complex Equation

**Viral** · October 10th 2009, 06:53 AM

$(3-i)(z+4-2i)=10+20i$

My attempt:

$\begin{array}{rcrcrc} (3-i)(z+4-2i)=10+20i\\ 3z + 12-6i-zi-4i+2i^{2}=10+20i\\ 3z+10-10i-zi=10+20i\\ 3z-zi=30i \end{array}$

I have no idea where to go from there.

**Defunkt** · October 10th 2009, 07:13 AM

Originally Posted by Viral

$(3-i)(z+4-2i)=10+20i$

My attempt:

$\begin{array}{rcrcrc} (3-i)(z+4-2i)=10+20i\\ 3z + 12-6i-zi-4i+2i^{2}=10+20i\\ 3z+10-10i-zi=10+20i\\ 3z-zi=30i \end{array}$

I have no idea where to go from there.

$3z - zi = 30i$

Let $z = x + yi \Rightarrow 3z - zi = 3(x+yi) - i(x+yi) = 3x + 3yi - xi + y = (3x + y) + (3y - x)i$

So:

$(3x+y) + (3y-x)i = 0 + 30i$

Can you solve from here?

**Viral** · October 10th 2009, 07:41 AM

I didn't understand how you worked the following out:

$3x + 3yi - xi + y = (3x + y) + (3y - x)i$

**Defunkt** · October 10th 2009, 07:47 AM

Originally Posted by Viral

I didn't understand how you worked the following out:

$3x + 3yi - xi + y = (3x + y) + (3y - x)i$

$3x + 3yi - xi + y = 3x+y+3yi-xi = 3x + y + i(3y) - i(x) = 3x + y + i(3y-x)$

**Viral** · October 10th 2009, 07:52 AM

Ehh, this is confusing. I've understood everything so far. So from that I'm guessing $3(x+y)=0, (3y+x)i=30i$ But I just don't see how I change that into $z=x+yi$ .

**Defunkt** · October 10th 2009, 07:57 AM

Originally Posted by Viral

Ehh, this is confusing. I've understood everything so far. So from that I'm guessing $3(x+y)=0, (3y+x)i=30i$ But I just don't see how I change that into $z=x+yi$ .

You are left with 2 equations that you need to solve for x,y:

$3x+y = 0 \Rightarrow x = \frac{-y}{3}$
$(3y-x)i = (30)i \Rightarrow 3y - x = 30$

Find the possible values for x,y then substitute them into $z = x + yi$ and get your values of z.

Spoiler:

**Viral** · October 10th 2009, 08:12 AM

How did you get $x=-y$ ?

**Plato** · October 10th 2009, 08:13 AM

Originally Posted by Viral

$(3-i)(z+4-2i)=10+20i$

You both are making this far too complicated.
Solve it like an elementary algebra problem: divide and then subtract.

$\frac{{10 + 20i}}{{3 - i}} = \frac{{(10 + 20i)(3 + i)}}{{10}} = 1 + 7i$ .

So $z=(1+7i)-(4-2i)$

**Viral** · October 10th 2009, 08:17 AM

$\frac{10+20i}{3-i}=(z+4-2i)$ .

But then how did you get 1+7i from there?

**Defunkt** · October 10th 2009, 08:24 AM

Originally Posted by Viral

$\frac{10+20i}{3-i}=(z+4-2i)$ .

But then how did you get 1+7i from there?

$\frac{10+20i}{3-i} = \frac{10+20i}{3-i} \cdot \frac{3+i}{3+i} =$ $\frac{(10+20i)(3+i)}{(3-i)(3+i)} = \frac{30 + 10i + 60i - 20}{3^2 + 1^2} =$ $\frac{10 + 70i}{10} = 1 + 7i$

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