A BIT COMPLEX. ISN'T IT?

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For complex numbers z and w, it is given that : ∣ z ∣ 2 w − ∣ w ∣ 2 z = z − w |z|^2w-|w|^2z=z-w . Then what is the value of w ‾ z \overline{w}z ?


The answer is 1.

Kirtan Bhatt by kirtan bhatt , 24, India

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3 solutions

D G
Dec 23, 2013

The question is badly formed.

Let z = a + b ∗ i z = a + b*i and w = c + d ∗ I w = c + d*I . Then the equation becomes:

( a 2 + b 2 ) ∗ ( c + d ∗ I ) − ( c 2 + d 2 ) ∗ ( a + b ∗ I ) = ( a + b ∗ I ) − ( c + d ∗ I ) (a^2 + b^2)*(c+d*I) - (c^2 + d^2)*(a + b*I) = (a + b*I) - (c + d*I)

Expanding:

c − a + i ∗ d − i ∗ b + c ∗ a 2 + c ∗ b 2 − a ∗ c 2 − a ∗ d 2 + i ∗ d ∗ a 2 + i ∗ d ∗ b 2 − i ∗ b ∗ c 2 − i ∗ b ∗ d 2 = 0 c - a + i*d - i*b + c*a^2 + c*b^2 - a*c^2 - a*d^2 + i*d*a^2 + i*d*b^2 - i*b*c^2 - i*b*d^2 = 0

i ∗ ( d − b + d ∗ a 2 + d ∗ b 2 − b ∗ c 2 − b ∗ d 2 ) + ( c − a + c ∗ a 2 + c ∗ b 2 − a ∗ c 2 − a ∗ d 2 ) = 0 i*(d - b + d*a^2 + d*b^2 - b*c^2 - b*d^2) + (c - a + c*a^2 + c*b^2 - a*c^2 - a*d^2) = 0

Let's substitute z = 1 + 1 ∗ i z = 1 + 1*i :

i ∗ ( − 1 + 3 ∗ d − c 2 − d 2 ) + ( − 1 + 3 ∗ c − c 2 − d 2 ) = 0 i*(-1 + 3*d - c^2 - d^2) + (-1 + 3*c - c^2 - d^2) = 0

Therefore:

( − 1 + 3 ∗ d − c 2 − d 2 ) = 0 (-1 + 3*d - c^2 - d^2) = 0 ( − 1 + 3 ∗ c − c 2 − d 2 ) = 0 (-1 + 3*c - c^2 - d^2) = 0

Solve this system of equations to yield c , d = 1 , 1 c, d = 1, 1 or c , d = 1 2 , 1 2 c, d = \frac{1}{2}, \frac{1}{2} .

c , d = 1 , 1 c, d = 1, 1 -> w ˉ z = 2 \bar{w}z = 2

c , d = 1 2 , 1 2 c, d = \frac{1}{2}, \frac{1}{2} -> w ˉ z = 1 \bar{w}z = 1

and therefore there is no unique solution.

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Harshil Patel
Dec 23, 2013

We can write ∣ z ∣ 2 = z z ˉ |z|^{2} = z\bar{z}

Rewriting above equation we get z z ˉ w − w w ˉ z = z − w z\bar{z}w - w\bar{w}z = z - w

= z z ˉ w − z + w − w w ˉ z z\bar{z}w - z + w - w\bar{w}z = z [ z ˉ w − 1 ] + w [ 1 − w ˉ z ] z[\bar{z}w-1] + w[1-\bar{w}z]

So this implies z [ z ˉ w − 1 ] = − w [ 1 − w ˉ z ] z[\bar{z}w-1] = - w[1-\bar{w}z]

Or z ˉ w − 1 \bar{z}w - 1 & 1 − w ˉ z = 0 1-\bar{w}z = 0

Implies z ˉ w \bar{z}w & w ˉ z \bar{w}z = 1

So w ˉ z \bar{w}z = * 1 \boxed{1} *

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Can you explain your 3rd last line?

Kulkul Chatterjee - 7 years, 5 months ago

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When we have A+B=0 we can say A=-B or these happens when both A and B are equal to 0

Harshil Patel - 7 years, 5 months ago

I am bound to say, that your 3rd last step is fatally incorrect. You remark that A=-B implies that A=B=0 ? Is that true? A=-B implies that A and B =0 if and only if A=B.

kirtan bhatt - 7 years, 5 months ago

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Well I said A+B = 0 implies A=-B or A+B = 0 can have a case where both A&B are equal to 0

Harshil Patel - 7 years, 5 months ago

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Do you think that this is the case in this problem?? How can you surely say that the assumption is correct??

kirtan bhatt - 7 years, 5 months ago

∣ z ∣ 2 w − ∣ w ∣ 2 z = z − w . H e n c e , w ( ∣ z ∣ 2 + 1 ) − z ( ∣ w ∣ 2 + 1 ) = 0. I t i m p l i e s t h a t , w z = ∣ w ∣ 2 + 1 ∣ z ∣ 2 + 1 . S o , i t s i m p l y m e a n s t h a t w z i s p u r e l y r e a l . N o w , s i n c e i t i s r e a l , w z = w ‾ z ‾ . H e n c e , z ‾ w = w ‾ z . N o w a g a i n , g o i n g b a c k t o t h e o r i g i n a l e q u a t i o i n , a n d p u t t i n g t h i s r e s u l t i n t o t h e e q u a t i o n , w e g e t , z w ‾ z − z + w − w w ‾ z = 0. O r , w ‾ z ( z − w ) − ( z − w ) = 0. H e n c e , ( w ‾ z − 1 ) ( z − w ) = 0. S o , z = w o r w ‾ z = 1. |z|^2w-|w|^2z=z-w. Hence, w(|z|^2+1)-z(|w|^2+1)=0. It implies that, \frac{w}{z}=\frac{|w|^2+1}{|z|^2+1}. So, it simply means that \frac{w}{z} is purely real. Now, since it is real, \frac{w}{z}=\frac{\overline{w}}{\overline{z}}. Hence, \overline{z}w=\overline{w}z. Now again, going back to the original equatioin, and putting this result into the equation, we get, z\overline{w}z-z+w-w\overline{w}z=0. Or, \overline{w}z(z-w)-(z-w)=0. Hence, (\overline{w}z-1)(z-w)=0. So, z=w or \overline{w}z=1.

kirtan bhatt - 7 years, 5 months ago
Eduardo Petry
Dec 31, 2013

∣ z ∣ 2 w − ∣ w ∣ 2 z = z − w → ( z w ) ( z − w ) = z − w → z w = 1 \displaystyle|z|^{2}w-|w|^{2}z=z-w \rightarrow (zw)(z-w)=z-w \rightarrow zw=\boxed{1}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

I know my resolution is wrong... I was actually lucky to find the right answer. I really need to learn more about complex numbers!

Eduardo Petry - 7 years, 5 months ago

my dear friend you cannot apply the general rules of factorization in this problem.... |z| and |w| denote the modulus of the complex number...

kirtan bhatt - 7 years, 4 months ago

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