Re: COmplex number uestion

what we are given:

|z| = 2√(10)

50/z* - 10/z = 2 + 9i

what i like to do is get everything over a common denominator:

50z/(zz*) - 10z*/(zz*) = 2 + 9i

50z - 10z* = (2 + 9i)(zz*)

now zz* = |z|^{2} = (2√(10))^{2} = 40 so we have:

50z - 10z* = (2 + 9i)(40)

take out a common factor of 10:

5z - z* = 8 + 36i

let's write this in terms of a and b, where z = a + bi

5(a + bi) - (a - bi) = 8 + 36i

(5a - a) + (5b + b)i = 8 + 36i

4a + 6bi = 8 + 36i

compare real and imaginary parts:

4a = 8 ---> a = 2

6b = 36 ---> b = 6.

so z = 2 + 6i. does this work?

first, we check |z| = √(a^{2} + b^{2}) = √(4 + 36) = √40 = √4√(10) = 2√(10). check.

next, let's find 1/z = 1/(2 + 6i) = (2 - 6i)/(4 + 36) = (1/40)(2 - 6i)

= (1/20)(1 - 3i).

finally, let's find 1/z* = z/|z|^{2} = (1/40)(2 + 6i) = (1/20)(1 + 3i).

so 50/z* = (50/20)(1 + 3i) = (5/2)(1 + 3i), and 10/z = (10/20)(1 - 3i) = (1/2)(1 - 3i).

subtract the second from the first: (5/2)(1 + 3i) - (1/2)(1 - 3i) =

(5/2 - 1/2) + (15/2 + 3/2)i = 4/2 + (18/2)i = 2 + 9i. check.

Re: COmplex number uestion

Thank you!

How did you get from

1/(2 + 6i) to (2 - 6i)/(4 + 36) ?

Thanks loads!

J.

Re: COmplex number uestion

Quote:

Originally Posted by

**Tutu** Thank you!

How did you get from

1/(2 + 6i) to (2 - 6i)/(4 + 36) ?

Thanks loads!

J.

Multiply top and bottom by the bottom's conjugate...

Re: COmplex number uestion

Quote:

Originally Posted by

**Tutu** Thank you!

How did you get from

1/(2 + 6i) to (2 - 6i)/(4 + 36) ?

Thanks loads!

J.

for any complex number a+bi:

1/(a+bi) = (a-bi)/(a^{2}+b^{2}).

how do we prove this?

suppose xy = 1. then y = 1/x. so if we want to show y = 1/x, we can instead show xy = 1.

so let's prove (a+bi)[(a-bi)/(a^{2}+b^{2})] = 1

the denominator in the second factor is just a real number, so we can "pull it out front":

(a+bi)[(a-bi)/(a^{2}+b^{2})] = (1/(a^{2}+b^{2}))[(a+bi)(a-bi)]

= (1/(a^{2}+b^{2}))(a^{2}-abi+abi-(bi)^{2})

= (1/(a^{2}+b^{2}))(a^{2}-b^{2}i^{2})

= ((1/(a^{2}+b^{2}))(a^{2}-b^{2}(-1))

= (1/(a^{2}+b^{2}))(a^{2}+b^{2}) = 1.

a shorter proof:

zz* = |z|^{2} so

zz*/|z|^{2} = 1, that is:

z(z*/|z|^{2}) = 1, so 1/z = z*/|z|^{2}.