Express 5√3-5i in polar form.
this expression is a bit ambiguous. Need to write it a bit better. Try the equation tool.
ok so \[5\sqrt{3} -5i\]
\[z=a+bi\] \[r=|z|=\sqrt{a^{2}+b^{2}}\] \[\theta=\tan^{-1}(\frac{ b }{ a })\] polar coordinates are in the form of \[(r,\theta)\]
so let \[z=5\sqrt{3}-5i\] a is the real part, y is the imaginary part so you have \[r=|z|=\sqrt{(5\sqrt{3})^{2}+(-5)^{2}}\]
you can finish the rest using this
I meant b is the imaginary part, not y, sorry
for some reason i got 40 for r, I don't know if its correct
you should get 25(3)+25=100 then sqrt 100= 10
if you write out your calculation step by step I can see where you went wrong
do I multiply the square root with 5 ?
yes
it is 5 lots of sqrt3 so 5sqrt3 you square it to get 5sqrt3 x 5sqrt3 which gives (5x5)(sqrt3 x sqrt3) which gives 25sqrt9 sqrt9=3 25(3)=75
ah, now I see, Thank you Very much
Continue through the question to make sure you get the correct final answer
for theta I got -60, but I think that wouldn't work.
you shouldn't get 60 at all
\[\theta = \tan^{1}(\frac{ b }{ a })= \tan^{-1}(\frac{ 5 }{ 5\sqrt{3} })\] |dw:1462032863617:dw|
you should get 30 degrees
1/6pi
sorry, I am still getting 60, I know your write because I know the answer but I don't know how
show me your working out
you are doing \[\tan^{-1}(\frac{ 5\sqrt{3} }{ 5 })\]
switch the fraction around
and you get 30
I am doing that,and still get 60. Maybe its the calculator
do \[\frac{ 5 }{ 5\sqrt{3} }\]
\[\tan^{-1}(ans)\]
man I'm still getting the answer. I don't want to take more of your time. Thanks!
wait maybe try \[5\sqrt{3}\] press equals then \[\frac{ 5 }{ ans }\] press equals then \[\tan^{-1}(ans)\] if this doesn't work then I give up xD
oh so now I got it thanks man!
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