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?

Hoare are you getting that $a^2 - b^2 + a^2 + b^2 = 4 -64$. I can't see why you are saying that at all. You aren't squaring both sides. You aren't multiplying both sides by the conjugate. You aren't squaring the square roots. What are you doing. — fleablood, Mar 05 '18 at 03:14
Please, remember that, in general, $(x+y)^2\ne x^2+y^2$. For instance, $(2+8i)^2=2^2+2\cdot2\cdot8i+(8i)^2=4+32i-64=-60+32i$. — egreg, Mar 05 '18 at 09:37
@DavidRicherby It definitely is the absolute value, as well as the modulus. Even for complex numbers, the two words are synonyms. — Misha Lavrov, Mar 05 '18 at 14:41
@MishaLavrov Oh, OK -- I've never that term used for complex numbers before. — David Richerby, Mar 05 '18 at 19:02

gt6989b · Accepted Answer · 2018-03-05T03:03:07.107

13

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.

edited Mar 05 '18 at 03:03

answered Mar 05 '18 at 02:45

gt6989b

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Thats how we are finding b right? There is other way than this? I think op was confused with a. – King Tut Mar 05 '18 at 02:52
@KingTut yes, but finding $a$ is easier of you know what $b$ is already :) – gt6989b Mar 05 '18 at 02:53
by using B as 8 I have gotten farther, but not by much. $$a^2+a\sqrt {a^2 + 64} = -30$$ squaring the radical doesn't help much and I am trying to find an efficient way to solve this. Unless that is the only way? – BouncySlime555 Mar 05 '18 at 02:53
@BouncySlime555 Before squaring, bring the radical to one side, and everything else to the other side. – saulspatz Mar 05 '18 at 03:02
@BouncySlime555 see update – gt6989b Mar 05 '18 at 03:03
Yes true gt69 liked the answer ! – King Tut Mar 05 '18 at 03:14

score 13 · Answer 2 · answered Mar 05 '18 at 03:24

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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: $\,z=2+8i-|z|=\ldots\,$

answered Mar 05 '18 at 03:24

dxiv

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This property strikes always – King Tut Mar 05 '18 at 03:58
@KingTut Many of these can be solved more directly without resorting to cartesian, indeed. – dxiv Mar 05 '18 at 04:02

score 7 · Answer 3 · answered Mar 05 '18 at 03:26

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

score 4 · Answer 4 · answered Mar 05 '18 at 02:38

4

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

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

answered Mar 05 '18 at 02:38

Siong Thye Goh

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score 2 · Answer 5 · answered Mar 05 '18 at 08:43

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

score 0 · Answer 6 · answered Mar 05 '18 at 17:40

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

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.

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

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

6 Answers6