## Destinyyyy one year ago perform the addition.. How do I solve this? The answer I got was incorrect.

1. Destinyyyy

(-5+5i) + (9+8i)

2. anonymous

4+13i

3. Destinyyyy

Yes I know that thats the correct answer. Im not asking for that. Im asking how to solve it

4. DecentNabeel

$\left(5i-5\right)+\left(8i+9\right)$ $\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$ $=\left(5i-5\right)+8i+9$ $\mathrm{Add/Subtract\:the\:numbers:}\:-5+9=4$ $=8i+5i+4$ $\mathrm{Add\:similar\:elements:}\:8i+5i=13i$ $=13i+4$ that is the answer

5. DecentNabeel

are you understand @Destinyyyy

6. Destinyyyy

Sorry one second while I read it.. There was a spider in my room :(

7. DecentNabeel

hahaha no problem @Destinyyyy

8. Destinyyyy

My example said to remove the i and put it outside of the equation.. (-5 +5) +(9+8)i

9. Destinyyyy

It was fluttering huge... Death by broom

10. Destinyyyy

it change f-ing to fluttering -.-

11. DecentNabeel

$\mathrm{Simplify}\:\left(-5+5\right):\quad 0$ $=i\left(8+9\right)+0$ $\mathrm{Simplify}\:i\left(8+9\right):\quad 17i$ 17i+0=17i

12. anonymous

The way your book does it is still true but i think its in this format (a+bi) + (d+ ci) = (a+d) (b+c) i

13. anonymous

To add or subtract we use Add (a+bi) + (c +di) = (a+b) + (c +d)i Subtract (a+bi) (c +di) = (a+b) - (c +d)i So in your case (-5+5i) + (9+8i) = (-5+5) + (9+8)i -5+9 + 5+8 4 + 13i

14. DecentNabeel

any confusion @Destinyyyy

15. DecentNabeel

@Deeezzzz -5+9=4 not -4

16. Destinyyyy

Okay.. I get it now.. I multiplied the parenthesis... Thank you everyone!!

17. anonymous

Makes sense?

18. Destinyyyy

Yes.. Can I get help with another one?

19. DecentNabeel

yes tell me @Destinyyyy

20. anonymous

Thanks! @DecentNabeel simple arithmetic mistake

21. DecentNabeel

no problem @Deeezzzz

22. Destinyyyy

Perform the addition (-9/2 +1/2i) + (3/2 - 7/2i)

23. DecentNabeel

$\left(-\frac{9}{2}+\frac{i}{2}\right)+\left(\frac{3}{2}-\frac{7i}{2}\right)$ $\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$ $=\left(-\frac{9}{2}+\frac{i}{2}\right)+\frac{3}{2}-\frac{7i}{2}$ $\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$ $=\frac{-7i+i+3-9}{2}$ refine $=\frac{-6i-6}{2}$ factor out -6 $=-\frac{6\left(i+1\right)}{2}$ $\mathrm{Divide\:the\:numbers:}\:\frac{6}{2}=3$ $=-\left(3\left(i+1\right)\right)$ $\mathrm{Negate}\:-\left(3\left(i+1\right)\right)=-3\left(i+1\right)$ $=-3\left(i+1\right)$ that is the answer

24. DecentNabeel

ok @Destinyyyy

25. Destinyyyy

??? Why is the 1 now i and the 7 now 7i?

26. Destinyyyy

My examples show to put the i on the outside of the equation like @Nixy did

27. DecentNabeel

-3i+3

28. Destinyyyy

Still confused

29. DecentNabeel

why confused tell me

30. DecentNabeel

you said i on the out sides .. so that.. 3(i+1) =3i+3

31. Destinyyyy

No I mean like this (-9/2 +1/2) + (3/2 - 7/2)i for the first step

32. DecentNabeel

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$ $=\left(-\frac{9}{2}+\frac{i}{2}\right)+\frac{3}{2}-\frac{7i}{2}$

33. DecentNabeel

alright @Destinyyyy

34. Destinyyyy

I dont understand why 1 is now i and why its 7i

35. DecentNabeel

ok i expla it

36. anonymous

$$\huge (-\frac{9}{2} +\frac{1}{2}i) + (\frac{3}{2} - \frac{7}{2}i)$$ Is the above how it looks in your book?

37. DecentNabeel

(-9/2 +1/2i) + (3/2 - 7/2i) that is your question

38. anonymous

$$\huge (-\frac{9}{2} +\frac{1}{2}i) + (\frac{3}{2} - \frac{7}{2}i)$$ $$\huge (-\frac{9}{2} +\frac{1}{2}) + (\frac{3}{2} - \frac{7}{2}) i$$ $$\huge (-\frac{9}{2} +\frac{3}{2}) + (\frac{1}{2} - \frac{7}{2}) i$$ $$\huge -\frac{6}{2} + (-\frac{6}{2}) i$$ $$\huge -3 -3i$$