Posted by **Tara** on Thursday, August 16, 2012 at 12:08pm.

I got this answer for a problem, and I'm not sure if it's right.

Can someone tell me if I'm correct?

1.(2+4i / 3-i)

= 6/7+5/7i

- Algebra2 -
**bobpursley**, Thursday, August 16, 2012 at 12:21pm
2+41 /(3-i)

rationalize

(2+4i)(3+i)/10= (6+2i+12i-4)/10=

=(2+14i)/10

- Algebra2 -
**Tara**, Thursday, August 16, 2012 at 12:31pm
Ahh!

I'm confused about this math stuff:/

it's my first time trying these algebra problems.

I'm trying to go by the teacher's examples.

Can You tell me if these are right:p?

2. (7-√-15)^2

= 34+14i√15i

3.4+√-20 / 2

=2+2.5√4i

- Algebra2 -
**bobpursley**, Thursday, August 16, 2012 at 12:32pm
another way, in polar form:

(2+4i)=sqrt20@arctan2

(3-i)=sqrt10@arctan(-1/3)

so (2+4i)/(3-i)=sqrt20/sqrt10 @ (arctan2-arctan-1/3)=sqrt2 @(1.11rad-(-.321))

= sqrt2 @1.43rad=sqrt2 @ 81.9 deg

Now converting this (only to compare with the above)

= sqrt2 (cos81.9+isin81.9)

= (.198+i*1.40)

and the above is .2+i1.40

I rounded the angles on the arctan conversions...

- Algebra2 -
**bobpursley**, Thursday, August 16, 2012 at 12:37pm
2. (7-√-15)^2

= 34+14i√15i ????

(7-isqrt15)^2=49-15-2isqrt15=34-2i sqrt15

3.4+√-20 / 2

=2+2.5√4i ????

4/2+1/2 i sqrt20=2+.5i*sqrt4*sqrt5

= 2+i sqrt5

- Algebra2 -
**Steve**, Thursday, August 16, 2012 at 1:16pm
(7-isqrt15)^2=49-15-2isqrt15??

I get 49-15- 14√15 i

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