How do you write the trigonometric form of #2+2i#?

Question

11559 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-25T12:07:13

The trigonometric form is #2sqrt2(cos(pi/4)+isin(pi/4))#

Let #z=2+2i#
To calculate the trigonomrtric version, we need to calculate the modulus of the complex number.

If #z=a+ib# then the modulus is #∣z∣=sqrt(a^2+b^2)#

So here #∣z∣=sqrt(2^2+2^2)=2sqrt2#
Then #z/(∣z∣)=1/sqrt2+(i)/sqrt2#

tHen we compare this to #z=r(costheta+isintheta)#

and we get #costheta=1/sqrt2#
and #sintheta=1/sqrt2#

so, #theta=pi/4#
and the trigonometric form is #2sqrt2(cos(pi/4)+isin(pi/4)#

And the exponential form is #z=2sqrt2e^(ipi/4)#