SAT Complex Numbers Problem

[Mampac]

Hello there, ran into this question, couldn't move further than ai - b = 1; according to the answers, the answer is 1. No explanation, however.

 

[Otis]

… ai - b = 1 …

Hello Mampac. That's a good start. Note that i appears on the left-hand side, but not on the right-hand side. Does that fact give you an idea about what must be the value of a?

?

 

[Steven G]

Take a+bi and multiply it by i. Then equate what you got to 1. This should tell you what a and b are so then you can compute a-b.

 

[pka]

View attachment 14786
Hello there, ran into this question, couldn't move further than ai - b = 1; according to the answers, the answer is 1. No explanation, however.

Notation: If \(\displaystyle z=a+bi\) then \(\displaystyle \Re(z)=a~\&~\Im(z)=b\) (the real & imaginary parts of \(\displaystyle z\))
So \(\displaystyle (a+bi)(i)=ai-b\) so that \(\displaystyle \Re(ai-b)=-b~\&~\Im(ai-b)=a\) but \(\displaystyle \Re(1)=1~\&~\Im(1)=0\)
BUT that means \(\displaystyle a=0~\&~b=-1\).

 

[Deleted member 4993]

View attachment 14786
Hello there, ran into this question, couldn't move further than ai - b = 1; according to the answers, the answer is 1. No explanation, however.

(a + bi) * i = 1

a*i - b = 1

a*i - b = 1 - 0*i

equating real terms and imaginary terms of the equation:

- b = 1 \(\displaystyle \ \to \ \) b = -1

a = 0

a - b = 0 - (-1) = 1 ................................edited (from the corner)

 

[Steven G]

(a + bi) * i = 1

a*i - b = 1

a*i - b = 1 - 0*i

equating real terms and imaginary terms of the equation:

- b = 1 \(\displaystyle \ \to \ \) b = -1

a = 0

a - b = 0 - (-1) = 0

Go directly to the corner for this error!

 

[Steven G]

Subhotosh meant to say that 0-(-1) =1

 

[Sonal7]

(a + bi) * i = 1

a*i - b = 1

a*i - b = 1 - 0*i

equating real terms and imaginary terms of the equation:

- b = 1 \(\displaystyle \ \to \ \) b = -1

a = 0

a - b = 0 - (-1) = 1 ................................edited (from the corner)

Why isnt there an i on the RHS when you multiplied the LHS with i? Thanks for the ans.

 

[Deleted member 4993]

Why isnt there an i on the RHS when you multiplied the LHS with i? Thanks for the ans.

I think you forgot what your original post was!!


I am starting with the first line of OP (question)!

Is it clear now?