A problem on complex no.

Question

Q.If $z$ and $w$ be two complex no, such that $|z|=|w|=1$ and$|z+iw|=|z-iw|=2$.Then find z

I tried to break $z$ and $w$ into $a+ib$ form separately and carry on linear equation by substituting the values.But,It leads to me nowhere.Any hints?

Answer 1

Interesting. I think "nowhere" is the right direction; I don't believe this has a solution.

I prefer to visualize this in terms of vectors and geometry on the complex plane. If z and w both have an absolute value of 1, they are both points on the unit circle.

Multiplying by $i$ is, in geometric terms, equivalent to rotating by 90 degrees counter-clockwise.

Translated geometrically, the last statement says "The z vector, added to a 90 degree counter-clockwise rotation of the w vector, ends up on the edge of a circle with radius 2. The z vector, added to a 90 degree clockwise rotation of the w vector, also ends up on the edge of a circle with radius 2."

Since both vectors have a magnitude of 1, they can only combine to 2 if they point in the same exact direction (after w's rotation). But the second sentence means they also get to the radius 2 circle if w points in the opposite direction. This is impossible unless w has 0 magnitude, but we already know it has a magnitude of 1. So it's impossible.

(Solving it algebraically winds up with a similar statement: $(expression) = -(expression) = 1$. Which is the essence of what I stated in my last paragraph.)

Answer 2

Let z = a + bi

Let w = c + di

iw = -d + ci

z + iw = (a-d) + i(b+c)

z - iw = (a+d) +i(b-c)

|z| = |w| = 1 => $a^2 + b^2 = c^2 + d^2 = 1$

|z + iw| =2 => $(a - d)^2 + (b + c)^2 = 4$ so $a^2 + b^2 + c^2 + d^2 - 2ad + 2bc =4$ so $2 + 2bc - 2ad =4$ so $bc - ad = 1$

Likewise |z - iw| =2 => $(a + d)^2 + (b - c)^2 = 4$ so $a^2 + b^2 + c^2 + d^2 + 2ad - 2bc =4$ so $2 + 2bc - 2ad =4$ so $ad - bc = 1$

So we have $bc - ad = 1$ and $ad - bc = 1$ which is impossible so, either it's late and I made a stupid error, or ... there's no solution.

Answer 3

It can be useful to get accustomed to not using the representation as $a+bi$.

The square of the norm of $z+iw$ is $(z+iw)(\bar{z}-i\bar{w})$. Multiplying, and using the fact that $z$ and $w$ have norm $1$, we get $$2+i(w\bar{z}-z\bar{w})=4.$$ Similarly, from the norm of $z-iw$ we get $$2-i(w\bar{z}-z\bar{w})=4.$$ This is impossible, for the two equations force $w\bar z-z\bar{w}=0$.