Divide a complex number

[Anthonyk2013]

I know this might be basic but I'm new to complex numbers.

Im trying to solve this problem.

U=2+5i

1+6i
U

1+6i • 2-5i
2+5i 2-5i =2-5i+12i+30
4-10i+10i+25 =2+30-5i+12i
4+25 =32+7i
29
Where have I gone wrong?

 

[evinda]

Your result is right!!! :wink:
I write it with Latex:
\(\displaystyle \frac{1+6i}{2+5i}=\frac{(1+6i)(2-5i)}{(2+5i)(2-5i)}=\frac{2-5i+12i+30}{4+25}=\frac{32+7i}{29}=\frac{32}{29}\)+\(\displaystyle \frac{7}{29}\)i

 

[Anthonyk2013]

Your result is right!!! :wink:
I write it with Latex:
\(\displaystyle \frac{1+6i}{2+5i}=\frac{(1+6i)(2-5i)}{(2+5i)(2-5i)}=\frac{2-5i+12i+30}{4+25}=\frac{32+7i}{29}=\frac{32}{29}+\frac{7}{29}i \)

What happens to 7i?

 

[evinda]

I edited my post..Do you understand it now or shall I explain it to you??

 

[pka]

I know this might be basic but I'm new to complex numbers.
Im trying to solve this problem.
U=2+5i
1+6i
U

One of the most useful and time-saving fact is \(\displaystyle \dfrac{1}{z} = \dfrac{{\overline z }}{{{{\left| z \right|}^2}}}\).
So in your question \(\displaystyle \dfrac{1}{{2 + 5i}} = \dfrac{{2 - 5i}}{{29}}\).

Now that saves a lot of work. All you now need is \(\displaystyle (1+6i)(2-5i)=32+7i\) divided ​29

 

[Anthonyk2013]

One of the most useful and time-saving fact is \(\displaystyle \dfrac{1}{z} = \dfrac{{\overline z }}{{{{\left| z \right|}^2}}}\).
So in your question \(\displaystyle \dfrac{1}{{2 + 5i}} = \dfrac{{2 - 5i}}{{29}}\).

Now that saves a lot of work. All you now need is \(\displaystyle (1+6i)(2-5i)=32+7i\) divided ​29

Thanks I understand now. Great help.

can you use that formula on all division

 

[pka]

dont understand where you got 29 from using that method

Well you no business trying this question if you don't know this.

\(\displaystyle \left| {a + bi} \right| = \sqrt {{a^2} + {b^2}} \)

So \(\displaystyle \left| {a + bi} \right|^2 = {{a^2} + {b^2}} \)

 

[Anthonyk2013]

Well you no business trying this question if you don't know this.

\(\displaystyle \left| {a + bi} \right| = \sqrt {{a^2} + {b^2}} \)

So \(\displaystyle \left| {a + bi} \right|^2 = {{a^2} + {b^2}} \)

I understand alright, just confused me for a moment. I'm new to this.