write principal value of Log[(1+√3i)^5] in form of a+ib
Answers
Answer:
First, know that each complex number can be written in the polar form
z=a+bi=reiφ,
for some r,φ∈R.
Now, eiφ=cosφ+isinφ. How this is done is a bit beyond the scope of this answer, but it can easily be seen from the Taylor series of ex, cosφ, and sinφ.
Notice that |cosφ+isinφ|=
√
cos2φ+sin2φ
=1, so r=|z|=
√
a2+b2
. Now, we have
a+bi=|z|eiφ=|z|(cosφ+isinφ)=|z|cosφ+|z|isinφ.
Equating the real and the imaginary parts, we see that
a=|z|cosφ,b=|z|sinφ,
i.e.,
φ=arccos
a
√
a2+b2
=arcsin
b
√
a2+b2
=argz.
This can also be seen if you draw a complex number of norm 1 in the complex coordinate system.
Once we have the polar form, computing logarithm is easy:
ln(a+bi)=ln(|z|eıφ)=ln|z|+lneıφ=ln|z|+ıφ.
Now, just substitute |z| and φ from the above, and you have the form you wanted
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