So, I'm having trouble with these complex equations. Any help would be great!

(2+3i)(3-4i)

2+i/3i

1+2i/1-i

Thanks again.

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- Nov 23rd 2009, 04:01 PMwafflemypandaComplex Equation Help!
So, I'm having trouble with these complex equations. Any help would be great!

(2+3i)(3-4i)

2+i/3i

1+2i/1-i

Thanks again. - Nov 23rd 2009, 04:13 PMqmech
Lets try something like your problems

(3+4i)(1+2i)

= 3*1 + 4i*1 + 3*2i + 4i*2i

= 3 + 4i + 6i + 8i^2

= 3 - 8 + 10i

= -5 + 10i.

See how it's done?

(3+4i) / ( 2+i )

= (3+4i)(2-i) / (2+i)(2-i) multiply by 1, where 1 = (2-i)/(2-i).

= multiply out the top / multiply out the bottom. - Nov 23rd 2009, 07:33 PMStroodle
Remember that $\displaystyle i^2=-1$ and if you multiply a complex number by its conjugate you will get a real number.

e.g. the conjugate of $\displaystyle a+bi$ is $\displaystyle a-bi$