Help
a) any attempts yet? as a reminder r = sqrt(x^2+y^2) and theta = arctan(y/x)
(17i)/13?
wiat
r = sqrt(13^2+17i^2)
we don't consider the i part (it is assumed that y is graphed in the imaginary plane) so sqrt(13^2+7^2) and arctan(7/13)
anyway, then you just need to simplify sqrt(13^2 + 7^2) to get sqrt(218) as the modulus z = __*[cos(theta) + isin(theta)] where the modulus goes into the blank and the theta value goes into the theta
z = _sqrt 218_*[cos(theta) + isin(theta)]
make sure to also plug in the theta value
1.43 righht
z = _sqrt 218_*[cos(1.43) + isin(1.43)]
arctan(7/13) = ?
z = _sqrt 218_*[cos(28.3) + isin(28.3)]
I would recommend using the radians value here since the other angle is in radians so 0.49
sqrt 218_*[cos(0.49) + isin(0.49)]
good then for b and c try to apply the multiplication and division rules of de moivres to your problem |dw:1527825303044:dw|
wouldnt it just be (13 + 7i) * 3(cos(1.43) + isin(1.43)).
notice how the numbers in front of your z and w roots are 3 and sqrt(218) so r1r2 = 3*sqrt(218)
notice also how it calls for theta1 + theta2 so you must add the theta value 0.49 obtained from step a) to 1.43 to obtain the new theta value
0.49+1.43
good so putting everything together what is zw?
1.92
3*sqrt(218)
that's the theta value, but what about the entire expression?
44.29
3*sqrt(218)=44.29
|dw:1527825826612:dw|
when you replace those values in, what do you get for the entire expression?
3sqrt(218)[cos(1.92)+isin(1.92)]
awesome so that's zw now repeat the same process for the division formula, be careful because order matters
is theta 1 0.49 or 1.43
it's z/w so theta1 will be the theta value for z (0.49)
(0.20)[cos(-0.94)+isin(-0.94)]
it's z/w so the r-value for z goes in the numerator sqrt(218)/3 should be the r value not the other way around
(4.9)[cos(-0.94)+isin(-0.94)]
good, so that's it
Join our real-time social learning platform and learn together with your friends!