How do you find the trigonometric form of the complex number 3i?

Question

How do you find the trigonometric form of the complex number 3i?

1 Answer

Impact of this question

8405 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-12-21T16:37:49

$z = 3 [cos (\frac{π}{2}) + i sin (\frac{π}{2})] = 3 i sin (\frac{π}{2})$ $z=3[cos(pi/2)+isin(pi/2)]=3isin(pi/2)$

When you have to convert a complex number, given in "rectangular form" ( $z = a + i b$ $z=a+ib$ ), to trigonometric form $z = r [cos (θ) + i sin (θ)]$ $z=r[cos(theta)+isin(theta)]$ you need to evaluate:
1) the modulus $r$ $r$ (using Pitagora's Theorem);
2) the argument $θ$ $theta$ (using trigonometry).

Graphically:
enter image source here

In your case you have: $z = 0 + 3 i = 3 i$ $z=0+3i=3i$ so that:
1) $r = \sqrt{3^{2} + 0^{2}} = 3$ $r=sqrt(3^2+0^2)=3$
2) $θ = arctan (\frac{3}{0}) = \frac{π}{2}$ $theta=arctan(3/0)=pi/2$

Graphically:
enter image source here

Science

Math

Humanities

... and beyond

How do you find the trigonometric form of the complex number 3i?

1 Answer

Related questions

Impact of this question