Adding Absolute value to a complex number: $ z+| z|=2+8i$

Question

I would like to know my error in this problem.

Find the complex number such that: $$ z+|z|=2+8i$$ So far, I have: $$ \begin{split} a+bi+\sqrt{a^2+b^2} &= 2 + 8i\\ a^2-b^2+a^2+b^2&=4-64\\ 2a^2 -b^2 + b^2&=-60\\ a^2&=-30 \end{split} $$

But I should end up with $$a^2=-15$$

No matter how hard I try, I can't seem to find what I did wrong. Any suggestions?

Answer 1

I would go about this differently. Since $|z| \in \mathbb{R}$, you know that $b=8$ immediately since $bi$ is the only imaginary term on the left and $8i$ - on the right.

Now the only thing is to find $a$...

UPDATE

We have the equation $$a + \sqrt{a^2+64} = 2$$ (hence $a<0$), which implies $$\sqrt{a^2+64} = 2-a$$ and now squaring will yield the desired result.

Answer 2

Yet another way: $\,z=2+8i-|z|\,$, so $\,\bar z =2-8i-|z|\,$, then multiplying the two:

$$\require{cancel} z \bar z = (2+8i-|z|)(2-8i-|z|) \;\iff\; \cancel{|z|^2} = \cancel{|z|^2} - 4|z| + 68 \;\iff\; |z| = 17 $$

Then, substituting back in the first equation: $\,z=2+8i-|z|=\ldots\,$

Answer 3

$a+bi+\sqrt{a^2+b^2} = 2 + 8i$ so

$a + \sqrt{a^2 + b^2} = 2$ and $b = 8$.

So $a + \sqrt{a^2 + 64} = 2$

So $\sqrt{a^2 + 64} = 2- a$

$a^2 + 64 = 4 -4a + a^2$

$4a = -60$

$a = -15$.

$z = -15 + 8i$.

....

To do what you were attempting

You have to realize that the $Re(z) = a + \sqrt{a^2 + b^2}$ and $Im(z) = b$. I think somehow you were thinking there were threee parts $Re(z)=a$ and $Im(z) = b$ and some $Weird(z) = \sqrt{a^2 + b^2}$ and that $z\overline z = Re^2(z) - Im^2(z) + Weird^2(z)$. That simply isn't true....

$(a+bi+\sqrt{a^2+b^2})(a - bi +\sqrt{a^2 + b^2}) = (2 + 8i)(2-8i)$

$(a + \sqrt{a^2 + b^2})^2 - b^2 = 4 - 64$

$2a^2 + b^2 + 2a\sqrt{a^2 + b^2} -b^2 = -60$

$a^2 + a \sqrt{a^2 + b^2} = -30$

which is a pain to solve but can be done.

Answer 4

As you square and take the real part, it should be

$$a^2-b^2+a^2+b^2+2a\sqrt{a^2+b^2}=4-64$$

Answer 5

For fun, solution using polar form. If $z = r e^{i\theta}$: $$re^{i\theta} + r = z + |z| = 2 + 8i,$$ $$re^{i\theta} = (2 - r) + 8i,$$ $$r = |re^{i\theta}| = |(2 - r) + 8i| = \sqrt{(2 - r)^2 + 8^2},$$ $$r^2 = (2 - r)^2 + 64,$$ $$r = 17,$$ $$z = re^{i\theta} = (2 - r) + 8i = -15 + 8i.$$

Answer 6

$$a+bi+\sqrt{a^2+b^2} = 2 + 8i$$ $$a^2-b^2+a^2+b^2=4-64$$

This is where you went wrong, it looks like you tried to square both sides and take the real part but you made the wrong assumption that all the cross-multiplication terms would be imaginary.

$$a+bi+\sqrt{a^2+b^2} = 2 + 8i$$ $$a^2-b^2+(a^2+b^2) + 2abi + 2a\sqrt{a^2+b^2} + 2b\sqrt{a^2+b^2}i = 4 - 64 +16i$$ $$a^2-b^2+a^2+b^2 + 2a\sqrt{a^2+b^2} = 4 - 64 $$ $$2ab + 2b\sqrt{a^2+b^2} = 16$$

Which frankly doesn't look much like progress towards a soloution. We have a pair of equations but both of them still contain our problematic square root.

To eliminate the square root we need to move it to it's own side of the equation before squaring.

$$a+bi+\sqrt{a^2+b^2} = 2 + 8i$$ $$\sqrt{a^2+b^2} = (2 - a)+ (8- b)i$$ $$a^2+b^2 =(2 - a)^2 - (8- b)^2 + 2(2 - a)(8- b)i $$ $$a^2+b^2 = a^2 - 4a + 4 - b^2 +8b -64 $$ $$2b^2 =- 4a + 4 +8b -60 $$ $$0 = (2 - a)(8- b)$$