Complex Numbers (2 Viewers)

[goobi]

This is a question from Patel Ex.4A.

Find x and y in (x-iy)^2=2i

Any help would be greatly appreciated!

 

[Alkanes]

x^2 - 2ixy - y^2 = 2i
x^2 - y^2 - i(2xy) = 2i

Equating to real and imaginary

x^2 - y^2 = 0
-2xy = 2

Solve =)

 

[SpiralFlex]

x^2 - 2ixy + y^2 = 2i
x^2 + y^2 - i(2xy) = 2i

Equating to real and imaginary

x^2 + y^2 = 0
2xy = 2

Solve =)

Good. Note: You must say equating real and imaginary coefficients otherwise you will not get the mark.

 

[Alkanes]

Lol sorry Spi. I have failed you.

 

[SpiralFlex]

Also it's,

for equating the imaginary part.

 

[Alkanes]

Ahh yes my bad.

 

[SpiralFlex]

Ahh yes my bad.

Also it's

 

[Alkanes]

Oh gee. I failed again.

 

[goobi]

x^2 - 2ixy + y^2 = 2i
x^2 + y^2 - i(2xy) = 2i

Equating to real and imaginary

x^2 - y^2 = 0
-2xy = 2

Solve =)

I actually did attempt to solve these simultaneous equations. However, I got 4 values of x which was weird...

Can you please help me solve them if you can?

Thank you (and also the robot) for the help btw!

 

[jnney]

yeah, you get (x^2-1) (x^2+1) = 0

but x^2 + 1 = 0 is not part of the solution for x because you are only finding REAL roots

therefore x = - and + 1

you then sub these values back into your original simultaneous equations (say, xy=-1) to find the values of y.

you get : (1, -1) and (-1, 1)

 

[Alkanes]

Ye i think you should get 2 values for x and 2 values for y.

Spi shall finish off the job, cos i might make a mistake again lols

 

[SpiralFlex]

If Goobi is confused,

Where





Equating real and imaginary coefficients,

(1)

(2)

From (2),



Into (1),









Where



Where

Remember cannot take imaginary solutions because it belongs into the real set only!

 

[artosis]

Oh have year 11's started 4 unit already!?
crap one year has gone by pretty quickly.

 

[Alkanes]

Lol time flies when you're having 'fun'

 

[goobi]

yeah, you get (x^2-1) (x^2+1) = 0

but x^2 + 1 = 0 is not part of the solution for x because you are only finding REAL roots

therefore x = - and + 1

you then sub these values back into your original simultaneous equations (say, xy=-1) to find the values of y.

you get : (1, -1) and (-1, 1)

Ye i think you should get 2 values for x and 2 values for y.

Spi shall finish off the job, cos i might make a mistake again lols

If Goobi is confused,

Where





Equating real and imaginary coefficients,

(1)

(2)

From (2),



Into (1),









Where



Where

Remember cannot take imaginary solutions because it belongs into the real set!

Got it. Thanks!

 

[jnney]

Oh have year 11's started 4 unit already!?
crap one year has gone by pretty quickly.

we are no longer in year 11.... >______<

 

[SpiralFlex]

Oh have year 11's started 4 unit already!?
crap one year has gone by pretty quickly.

Hence why I believe 2012 cohort is going to be the best yet. Keen students.

Got it. Thanks!

Let me eat you now.

 

[Alkanes]

The feeling of yr 12 hasn't sunked in yet. I'll still consider myself a yr 11 until 2012 starts =)

 

[goobi]

Let me eat you now.

Before eating me, would you like to have some Dim Sims as your first course?

 

[kayven]

Oh My Gosh man. You guys are so smart at Maths, im in year 11 as well and i just tried and self teach myself basics of Complex Number. :'( Completing against you guys in Extension 2 Maths gives me shiver.
:O

Also, how much hours do you guys study daily?