How do you write the complex number in trigonometric form #-4+4i#?

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Answer 1 · 2016-11-26T15:47:06

The answer is #=4sqrt2(cos((3pi)/4)+isin((3pi)/4))#

Let #z=a+ib# be a complex number.

To convert to trigonometric form

#z=r(costheta+isintheta)#

We calculate the modulus, #∥z∥=sqrt(a^2+b^2)#

#z=-4+4i#

#∥z∥=sqrt(16+16)=sqrt32=4sqrt2#

#z=4sqrt2(-4/(4sqrt2)+(4i)/(4sqrt2))#

#=4sqrt2(-1/sqrt2+i/sqrt2)#

So, #r=4sqrt2#

#costheta=-1/sqrt2#

#sintheta=1/sqrt2#

We are in the second quadrant

#theta=(3pi)/4#

#z=4sqrt2(cos((3pi)/4)+isin((3pi)/4))#