What is the trigonometric form of # 12+4i #?

2 Answers
Shwetank Mauria
May 19, 2016

#12+4i=4sqrt10(cosalpha+isinalpha)#, where #tanalpha=1/3#

Explanation:

#a+ib# can be written in trigonometric form #re^(itheta)=rcostheta+irsintheta=r(costheta+isintheta)#,
where #r=sqrt(a^2+b^2)#.

Hence #12+4i=sqrt(12^2+4^2)[cosalpha+isinalpha]#

= #sqrt160[cosalpha+isinalpha]#

= #4sqrt10[cosalpha+isinalpha]#,

where #cosalpha=12/(4sqrt10)=3/sqrt10#

and #sinalpha=4/(4sqrt10)=1/sqrt10#

or #tanalpha=(1/sqrt10)/(3/sqrt10)=1/3#

Jim G.
May 19, 2016

#4sqrt10(cos(0.322)+isin(0.322))#

Explanation:

Given a complex number z = x + yi , this can be written in trig. form as.

#z=x+yi=r(costheta+isintheta)#

where #r=sqrt(x^2+y^2)#

and #theta=tan^-1(y/x)#

here x = 12 and y = 4

#rArrr=sqrt(12^2+4^2)=sqrt160=4sqrt10#

and #theta=tan^-1(4/12)≈0.322" radians or"18.43^@#

#rArr12+4i=4sqrt10(cos(0.322)+isin(0.322))#or

#12+4i=4sqrt10(cos(18.43)^@+isin(18.43)^@)#

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