How do you add $(2 + 9 i) + (5 - 7 i)$ $(2+9i)+(5-7i)$ in trigonometric form?

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Answer 1 · 2018-05-21T14:58:20

$(2 + 9 i) + (5 - 7 i) = \sqrt{53} (cos (0.279) - sin (0.279) i)$ $(2+9i)+(5-7i)=sqrt53(cos(0.279)-sin(0.279)i)$

We add two complex numbers $a + b i$ $a+bi$ and $c + d i$ $c+di$ as follows:

$(a+bi)+(c+di):=(a+c)+(b+d)i$

So $(2+9i)+(5-7i)=7-2i$ . Now, to convert to trigonometric, we use the following identity:

$(a+bi)=r(cos(theta)+sin(theta)i)$

where $r=sqrt(a^2+b^2)$ and $theta$ satisifies $costheta=a/r,sintheta=b/r$

So, for $7-2i$ , $r =sqrt(49+4)=sqrt53$ and $costheta=7"/"sqrt53,sintheta=-2"/"sqrt53$ and so $theta=arctan(-2"/"7)approx-0.279$

Finally, $7-2i=sqrt53(cos(-0.279)+sin(-0.279)i)=sqrt53(cos(0.279)-sin(0.279)i)$

Answer 2 · 2018-07-09T11:53:02

$color(violet)(=> 7 + 2 i$

$z= a+bi= r (costheta+isintheta)$

$r=sqrt(a^2+b^2), " " theta=tan^-1(b/a)$

$r_1(cos(theta_1)+isin(theta_2))+r_2(cos(theta_2)+isin(theta_2))=r_1cos(theta_1)+r_2cos(theta_2)+i(r_1sin(theta_1)+r_2sin(theta_2))$

$r_1=sqrt(2^2+ 9^2))=sqrt 85$
$r_2=sqrt(5^2+ -7^2) =sqrt 74$

$theta_1=tan^-1(9 / 2)~~ 77.47^@, " I quadrant"$
$theta_2=tan^-1(-7/ 5)~~ 305.54^@, " IV quadrant"$

$z_1 + z_2 = sqrt 85 cos(77.47) + sqrt 74 cos(305.54) + i (sqrt 85 sin 77.47 + sqrt 74 sin 305.54)$

$=> 2 + 5 + i (9 - 7 )$

$color(violet)(=> 7 + 2 i$

How do you add (2+9i)+(5-7i)(2+9i)+(5−7i)(2+9i)+(5-7i) in trigonometric form?