helllllllllllp plsssssss The complex number z is given by z = (√3) + i. Showing your working, express in the form x + iy, where x and y are real, (a) iz*/z
@perl
@nincompoop
do you know what z* means or is?
conjugate
@phi
write down what you have so far
first tell me should we multiply like this i(\[\sqrt{3}-i\] )= sqrt3i -i^2
yes, that is correct. however, remember i*i is -1
yes but am stuck.. am not getting the right answer -_-
could solve pls
your not stuck. just not finished. so fare you have \[ \frac{i \ z*}{z} = \frac{1+ \sqrt{3} \ i}{\sqrt{3} + i} \]
to get rid of a complex number in the denominator, multiply it by its complex conjugate. to keep things equal, do the same to the top
i already done all this... am stuck.. i think i have not done it right
let me try again
first, what do you get for \[ (\sqrt{3} + i)(\sqrt{3} - i)\]?
its the signs problem
do you know FOIL? for multiplying
no
FOIL is a way to remember what to multiply \[ (\sqrt{3} + i)(\sqrt{3} - i) \] F irst \( \sqrt{3} \sqrt{3} \) Outer \( \sqrt{3} \cdot -i \) Inner \( i \cdot \sqrt{3} \) Last \( i \cdot -i\)
the other way to remember is use the distributive law
am having a problem with the numerator
First, what do you get for the bottom ?
\[(1+\sqrt{3}i) \times (\sqrt{3}-i)\]
for denominator ive got 4
ok. now the top. You can use FOIL \[ (1+\sqrt{3}i) (\sqrt{3}-i) \] First \( 1 \cdot \sqrt{3} \) Outer \( 1 \cdot -i\) Inner \( \sqrt{3} i \cdot \sqrt{3} \) Last \( \sqrt{3} i \cdot -i\)
then what should i get ?
simplify each term. Expect 2 reals (no i) and 2 imag (with an i) the first two F and O are easy to simplify
sorry ?
yeah i have understood and got
the F (i.e. First) term is 1 * sqr(3) which simplifies to sqr(3)
yes
now simplify the next term (the Outer)
-i
ok, now the Inner term
3i
and the Last
sqrt 3
now add them up and "combine like terms" \[ \sqrt{3} -i+3i +\sqrt{3} \]
the answer is right now.. thank you so much.. i have another problem
please make it a new post. This one is a bit long.
yes
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