Complex Numbers

Hello!

This question is in the complex numbers chapter of FP1 and is in the section where you are taught how to divide by complex numbers:

Find real numbers a and b with a>0 such that:
(a+bj)^2 = 21 + 20j

There is probably a very simple method to solve this but I have spent about half am hour trying to figure it out and I've got myself more confused than I started!

Thank you for any help :smile:

Reply 1

Kevin De Bruyne

21

Original post by amydowse

Hello!

This question is in the complex numbers chapter of FP1 and is in the section where you are taught how to divide by complex numbers:

Find real numbers a and b with a>0 such that:
(a+bj)^2 = 21 + 20j

There is probably a very simple method to solve this but I have spent about half am hour trying to figure it out and I've got myself more confused than I started!

Thank you for any help :smile:

How far did you get?

I would suggest expanding the LBS and equating the real and imaginary parts of the equations.

Reply 2

Doctor_Einstein

19

Original post by amydowse

Hello!

This question is in the complex numbers chapter of FP1 and is in the section where you are taught how to divide by complex numbers:

Find real numbers a and b with a>0 such that:
(a+bj)^2 = 21 + 20j

There is probably a very simple method to solve this but I have spent about half am hour trying to figure it out and I've got myself more confused than I started!

Thank you for any help :smile:

Step 1. Expand the brackets

Step 2. Simplify by setting the j^2 = -1

Step 3. Real part of LHS = 21, Complex part of LHS = 20

Step 4. Using the two simultaneous from step 3, solve for a, and b

If you are still stuck, post your attempt at some of these steps here.

(edited 9 years ago)

Reply 3

username1732133

15

You can also use the modulus and argument.

If you're feeling lazy just put do this..

a=(21^2+20^2)^{1/4}\left( \cos\left( \frac{1}{2} \arctan(\frac{20}{21}\right) \right)

Reply 4

amydowse

OP

1

Original post by Doctor_Einstein

Step 1. Expand the brackets

Step 2. Simplify by setting the j^2 = -1

Step 3. Real part of LHS = 21, Complex part of LHS = 20

Step 4. Using the two simultaneous from step 3, solve for a, and b

If you are still stuck, post your attempt at some of these steps here.

Thank you - I've managed up to step 3 but when I try and solve the simultaneous equations I'm not getting the right answers!

Reply 5

Doctor_Einstein

19

Original post by amydowse

Thank you - I've managed up to step 3 but when I try and solve the simultaneous equations I'm not getting the right answers!

Good work so far.

Now express b = 10/a

Now substitute this into the equation a^2 - b^2 = 21, to yield:

a^2 - 100/a^2 = 21

Therefore,

a^4 - 100 = 21a^2

a^4 - 21a^2 - 100 = 0

This is a quadratic equation. To show this, let p = a^2, and we arrive at:

p^2 - 21p - 100 = 0

Solve for p, thus find a, and thus find b=10/a