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How do you simplify #(4-6i)(2+3i)#?

Answer 1

#(4-6i)(2+3i)= 26#

#{: (xx,color(white)("X")"|",4,-6i,), ("----",,"----","----",), (color(white)("X")2,color(white)("X")"|",color(white)("X")color(red)(8),color(green)(-12i),), (+3i,color(white)("X")"|",color(green)(12i),color(blue)(+18),), ("----",,"----","----",), (color(red)(8),color(green)(+0i),color(blue)(+18),,=26) :}#

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Answer 2

Cora Alexander

Given: #(4-6i)color(brown)((3+3i))#

#4color(brown)((2+3i)) -6icolor(brown)((2+3i))#...........................(1)

#(8+12i) + (-12i-18(i)^2)#...................(2)

Consider #18i^2#

#i^2 =-1#

So #18i^2= -18#

Thus #-18i^2= -(-18)=+18#....................(3)

Substitute (3) into (2) giving

#(8+12i)+(-12i +18)#

#8 +(12i -12i) + 18#

#8+0+18 = 26#

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Answer 3

Elias Ackerman

To simplify the expression ((4-6i)(2+3i)), you can use the distributive property of multiplication over addition:

[ (a + bi)(c + di) = ac + adi + bci + bdi^2 ]

Substituting the given values:

[ (4 - 6i)(2 + 3i) = 4(2) + 4(3i) - 6i(2) - 6i(3i) ]

[ = 8 + 12i - 12i - 18i^2 ]

Recall that ( i^2 = -1 ), so substituting:

[ = 8 + 12i - 12i - 18(-1) ]

[ = 8 + 12i - 12i + 18 ]

[ = 26 ]

So, ((4-6i)(2+3i)) simplifies to (26).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.