Problem 136
Easy Powers and Roots

Write Ln(2-3i) in the form u+iv

Write $\text{Ln}(2-3i)$ in the form $u+iv$.

Explanation of Method

The principal logarithm is $\text{Ln}(z)=\ln|z|+i\,\text{Arg}(z)$, with $\text{Arg}(z)\in(-\pi,\pi]$.

Step 1 · Modulus

$$ |2-3i|=\sqrt{2^2+(-3)^2}=\sqrt{13} $$

Step 2 · Principal argument

Since $2-3i$ lies in the fourth quadrant ($x=2>0,\ y=-3<0$), the principal argument is given directly by $\arctan(y/x)$:

$$ \text{Arg}(2-3i)=\arctan\left(\frac{-3}2\right)=-\arctan\frac32 $$

Step 3 · Combine

$$ \text{Ln}(2-3i)=\ln\sqrt{13}-i\arctan\frac32=\frac12\ln13-i\arctan\frac32 $$

Step 4 · Numerical values

$$ u=\frac12\ln13\approx1.282475, \qquad v=-\arctan\frac32\approx-0.982794\text{ rad}\ (\approx-56.31^\circ) $$

Result

$$ \boxed{\text{Ln}(2-3i)=\frac12\ln13-i\arctan\frac32\approx1.2825-0.9828i} $$

Since $2-3i$ already sits in the fourth quadrant, its principal argument needs no quadrant adjustment — it's simply $\arctan(-3/2)$.