Question

(a+bi)^2 = -2+5i; a,b>o

Answer 1

Again, foil it out. If you see an \(i^2\) swap it with a (-1)

Answer 2

\[(a+bi)^{2} =-2+5i\]

Answer 3

\[(a + bi)(a+bi) \]\[= a^2 + 2abi + b^2i^2 \]\[=(a^2 - b^2) + (2ab)i = -2+5i\]
\[\implies \begin{array}{ccc} a^2 - b^2 = -2 & and & 2ab = 5\end{array}\]

Answer 4

Does this now become some sort of simultaneous equation?

Answer 5

It is a system of equations that you'll have to solve for a and b.

Answer 6

I suggest the substitution method..

Answer 7

\[b = \frac{5}{2a} \implies b^2 = \frac{25}{4a^2}\]

Answer 8

You're a legend. Thanks a lot! A lot easier then I was told in class.

Answer 9

How was it explained in class?

Answer 10

The only real trick is to foil, then equate the coefficients of the real and imaginary parts.

Answer 11

Is this algebra 2?

Answer 12

It was explained VERY differently. Using the quadratic formula, amung other things.

I don't know - we call it Maths C in Australia. Complex numbers, as you may have guessed.

Answer 13

Well you will need the quadratic formula

Answer 14

Once you substitute in the \(b^2\) term

Answer 15

Cool. Thanks sir.

Answer 16

Oh, an aussi eh? We don't get many of them around here.

Answer 17

This is for pre-university though yes?

Answer 18

I guess so. I'm in year 11. I go to university after next year. Why do you ask?

Answer 19

Just curious

Answer 20

May I ask your profession?

Answer 21

I'm a student at university. I also do some programming and sysadmin work on the side.

Answer 22

That's good. :)

Answer 23

So did you get this system solved yet?

Answer 24

Or are you still working on it?

Answer 25

a^2 -(25/4a^2) = -2

Multiply every term by 4a^2 to get rid of the denominator in the fraction.

4a^4 - 25 = -8a^2
4a^4 + 8a^2 -25 = 0

4x^2 + 8x - 25 = 0 where x = a^2
x = -3.7
x = 1.7

x = a^2
a = sqrt(-3.7)
b = sqrt(1.7)

a = 1.9i
a = -1.9i
b = 1.3
b = -1.3

Does this seem about right? Thanks in advance.

Answer 26

Recall however that the coefficients a and b are both real, and they are both given to be positive.

Answer 27

So the only valid solution for a is sqrt(1.7)

Answer 28

And that would make
\[b = \frac{5}{2\sqrt{1.7}}\]

Answer 29

Aaaah; because the restrictions on the equation at the start where a,b > 0?
So the real answers would be:
a = 1.9i
b = 1.3

Correct?

Answer 30

a cannot be 1.9i

Answer 31

I think you are confusing yourself. back up to the solutions you got for x.

Answer 32

x = a^2
a = sqrt(-3.7)
b = sqrt(1.7)

a = 1.9i
b = 1.3

Square root the negative number you get a result which is i?

Answer 33

How would you go about solving this equation?

4a^4 + 8a^2 -25 = 0

Answer 34

\[x \in \{-3.7,1.7\}\]\[\implies a^2 \in \{-3.7,1.7\}\]

Answer 35

You solved it the right way. You just miss-interpreted the results.

Answer 36

We get two different solutions for a^2

Answer 37

-3.7 and 1.7

Answer 38

But a must be real (its the coefficient)

Answer 39

So the square of a cannot be negative.

Answer 40

So we can discard the -3.7 result. and we are left with \[a = \pm \sqrt{1.7}\]

Answer 41

But since we are given that a must be positive we can throw away the negative square root also.

Answer 42

So for sure \[a = \sqrt{1.7} \approx 1.303\]

Answer 43

Now we just go back to 2ab = 5 to solve for b.

Answer 44

The coefficients of complex numbers are always real.

Answer 45

Aaah okay. Thanks so much.

We can actually leave it as a^2-b^2 = 2

So if a^2 = 1.7
a^2-b^2 = 1.7
1.7 - b^2 = 2
-b^2 = 0.3
-b = sqrt(0.3)
b = -0.54?

Or must you use the other equation? the second equation?

Answer 46

That's not right.

Answer 47

\[a^2 - b^2 = -2\]\[\implies 1.7 - b^2 = -2 \]\[\implies -b^2 = -3.7\]\[ \implies b^2 = 3.7 \implies b = \sqrt{3.7}\]

Answer 48

Cool. That makes sense! Thank you so much! You time is truely appreciated!

Answer 49

Sure thing =)

Answer 50

Surely you have something better to do with your time?

Answer 51

At this moment, it's either this, or sleeping. I'll go with this for a while.