Subtract the complex numbers: (4 + 3i) - (6 - 2i)
The difference of the two complex numbers (a+bi) and (c+di) , is --> ((a-c) + (b-d)i) ( 4 + 3i) - ( 6 - 2i) = (-2 + 5i)
\(\color{#0cbb34}{\text{Originally Posted by}}\) @XioGonz The difference of the two complex numbers (a+bi) and (c+di) , is --> ((a-c) + (b-d)i) ( 4 + 3i) - ( 6 - 2i) = (-2 + 5i) \(\color{#0cbb34}{\text{End of Quote}}\) Multiply the complex numbers: (2 + i)(3 - 6i) *
If you'd like you can make a new post and tag me there if you want more help (:
\(\color{#0cbb34}{\text{Originally Posted by}}\) @Zelynross \(\color{#0cbb34}{\text{Originally Posted by}}\) @XioGonz The difference of the two complex numbers (a+bi) and (c+di) , is --> ((a-c) + (b-d)i) ( 4 + 3i) - ( 6 - 2i) = (-2 + 5i) \(\color{#0cbb34}{\text{End of Quote}}\) Multiply the complex numbers: (2 + i)(3 - 6i) * \(\color{#0cbb34}{\text{End of Quote}}\) It’s basically the same jus multiplying (A+c)(b-d)(i) (2 + i)(3 - 6i) = (6 + -5)
\(\color{#0cbb34}{\text{Originally Posted by}}\) @CripQUEZZ \(\color{#0cbb34}{\text{Originally Posted by}}\) @Zelynross \(\color{#0cbb34}{\text{Originally Posted by}}\) @XioGonz The difference of the two complex numbers (a+bi) and (c+di) , is --> ((a-c) + (b-d)i) ( 4 + 3i) - ( 6 - 2i) = (-2 + 5i) \(\color{#0cbb34}{\text{End of Quote}}\) Multiply the complex numbers: (2 + i)(3 - 6i) * \(\color{#0cbb34}{\text{End of Quote}}\) It’s basically the same jus multiplying (A+c)(b-d)(i) (2 + i)(3 - 6i) = (6 + -5) \(\color{#0cbb34}{\text{End of Quote}}\) Whoops I meant (12 - 9i)
after all of my very hard research non plagirized asnwers id say 69has to be the corrects answer
\(\color{#0cbb34}{\text{Originally Posted by}}\) @elpapitocay after all of my very hard research non plagirized asnwers id say 69has to be the corrects answer \(\color{#0cbb34}{\text{End of Quote}}\) After all my very hard research I’ve come to the conclusion that u are slow
what?
you dnt knon wat to say cause yuh scared yuh blocke meh for no reason and imma unfollow you on brainly
homie
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