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Find real number $a$ such that a complex number $z$ may satisfy the equation

$|z|^2 -2zi +2a(1+i)=0$

Where $i=\sqrt {-1}$

I came across this problem while attempting my teachers homework and made attempts to solve this. I assumed $z$ to be $p+iq$ and reached the conclusion that point $z$ is at a distance of $\sqrt{2}$ from the point $-1-i$ on the Argand plane. However i am unable to proceed from here.

Any help to solve this problem would be appreciated. Thanks in advance.

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  • $\begingroup$ 1. Perhaps show how you obtained the $\sqrt2$ distance. 2, "Find real number 'a' such that": haven you copied/read the question accurately? $\endgroup$
    – ryang
    Commented Nov 12, 2021 at 3:08

3 Answers 3

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Assuming that $z = (x + iy)$, the imaginary portion of the equation

$$|z|^2 - 2zi + 2a(1 + i) = 0$$

becomes $-2ix + 2ia = 0 \iff a = x$.

Therefore, requiring that $z$ have form $(a + iy)$ is both necessary and sufficient with respect to the imaginary portion of the equation.

Therefore, all values of $(a)$ must be determined so that for each such value, at least one real value for $y$ exists so that

$$(z = a + iy) ~~\text{and} ~~|z|^2 - 2zi + 2a(1 + i) = 0.\tag1$$

Since the imaginary portion of (1) above is now guaranteed to be satisfied, attention may be confined to the real portion of (1) above.


First of all, it is immediately obvious that $a = 0$ is viable, as it leads to $z = 0$, for which (1) above is satisfied.

Therefore, without loss of generality, $a \neq 0$.
Under this assumption, the math is simplified by assuming that $y = ak ~ \implies z = a(1 + ik).$

Then $|z|^2 = a^2 (1 + k^2)$.

Therefore, the real portion of (1) above is transformed to

$$a^2(1 + k^2) + 2ak + 2a = 0 \iff a^2(1 + k^2) + 2a(1 + k) = 0. $$

Since it is assumed that $a \neq 0$, the above equation can be transformed to

$$a(1 + k^2) + 2k + 2 = 0 \iff k^2(a) + k(2) + (a + 2) = 0.\tag2 $$


Therefore, the entire problem has been reduced to identifying for which values of $a$ (2) above is solvable, where $k$ is required to be a real number. It is immediate that $k$ is solvable if and only if the discriminant to (2) above is $\geq 0$.

Therefore, the problem reduces to identifying for which values of $a \neq 0$, you have that

$$\left(4 - 4a^2 - 8a\right) \geq 0 \iff (a^2 + 2a - 1) \leq 0.$$

$$a^2 + 2a -1 = 0 \implies a = \frac{1}{2}\left[-2 \pm \sqrt{8}\right] = -1 \pm \sqrt{2}.\tag3$$

Therefore, based on (3) above, for $a \neq 0$, (2) above is solvable if and only if

$$\left( ~-1 -\sqrt{2} ~\right) ~\leq ~a ~\leq ~\left( ~-1 + \sqrt{2} ~\right).\tag4$$

Since the range given in (4) above includes $a = 0$, the range given in (4) above is the final answer.

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When you expand out your equation, you can write it in terms of its real and imaginary parts. Remember that $0 = 0 + 0i$, so the real part of the left-hand side must equal zero (giving you one equation in terms of $x$, $y$ and $a$) and so must the right-hand side (giving you a second equation in terms of $x$, $y$, and $a$).

That's two equations in two unknowns, but the real part of the equation actually has another important property - you can write it as a sum of squares and a residual, in a way that offers at least one obvious choice of $x$ and $y$, which then leads to the desired value of $a$.

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  • $\begingroup$ I tried this before and got x=a...and couldnt proceed with the 1st equation.do u want me to assume x &y as per my choice? $\endgroup$ Commented Nov 12, 2021 at 3:14
  • $\begingroup$ Assuming the problem is as you've written, then you just need to find one value of $a$ that will let you get a solution. If you put $x = a$ into the real part, then you get an equation in $x$ and $y$ that very strongly hints at values to make them work. $\endgroup$
    – ConMan
    Commented Nov 12, 2021 at 3:22
  • $\begingroup$ So is it just like i am free to put any value of x&y as per my choice for the sake of getting the ans?But if i assume x then am i not indirectly assuming a??then does it mean the problem is incomplete? $\endgroup$ Commented Nov 12, 2021 at 3:34
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Pair the equation with its complex conjugate:

$|z|^2-2zi+2a(1+i)=0$

$|z|^2+2\overline zi+2a(1-i)=0$

Subtracting the conjugated equation from the original gives a linear relation between $z$ and $\overline z$:

$-2i(z+\overline z)+4ai=0$

Thus $z+\overline z=2a$ forcing $z=a+yi$ for some real number $y$. Note that this implies that $a$ is just the $x$ coordinate on the locus you found (a circle with radius $\sqrt2$ centered at $(-1,-1)$), so if you have that locus in hand you should immediately be able to read off the domain.

Here I continue with the algebraic solution. Plug the known form $z=a+yi$ into the original equation:

$a^2+y^2-2ai+2y+2a(1+i)=0$

$y^2+2y+(a^2+2a)=0$

This has a real root for $y$, as required, iff the discriminant $4-4(a^2+2a)\ge0$, thus $a^2+2a-1\le 0$. So $a$ lies inclusively between the zeroes of $a^2+2a-1$:

$-1-\sqrt2\le a\le -1+\sqrt2$

which, as expected, matches the domain 9f $x$ coordinates for your circle.

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