Problem 159
Easy Limits and Functions

Find the real part u and imaginary part v of w=z^2e^(2z)

Find the real part $u$ and imaginary part $v$ of $\ w=z^2e^{2z}$.

Explanation of Method

Expand $z^2$ and $e^{2z}$ separately into real/imaginary form, then multiply the two complex factors together.

Step 1 · Expand $z^2$

$$ z^2=(x+iy)^2=(x^2-y^2)+i(2xy) $$

Step 2 · Expand $e^{2z}$

$$ e^{2z}=e^{2x}e^{2iy}=e^{2x}[\cos(2y)+i\sin(2y)] $$

Step 3 · Multiply

$$ w=e^{2x}\big[(x^2-y^2)+i(2xy)\big]\big[\cos(2y)+i\sin(2y)\big] $$

Expanding the product of the two complex factors ($(a+ib)(c+id)=(ac-bd)+i(ad+bc)$ with $a=x^2-y^2,\ b=2xy,\ c=\cos2y,\ d=\sin2y$):

$$ w=e^{2x}\Big[\big((x^2-y^2)\cos2y-2xy\sin2y\big)+i\big((x^2-y^2)\sin2y+2xy\cos2y\big)\Big] $$

Step 4 · Read off $u,v$

$$ \boxed{u(x,y)=e^{2x}\big[(x^2-y^2)\cos2y-2xy\sin2y\big], \qquad v(x,y)=e^{2x}\big[(x^2-y^2)\sin2y+2xy\cos2y\big]} $$

Verification, $z=0$

$u(0,0)=e^0[0-0]=0,\ v(0,0)=e^0[0+0]=0$; direct check $0^2e^0=0$ ✓.

Verification, $z=1$ (real)

$u(1,0)=e^2[(1)\cdot1-0]=e^2,\ v(1,0)=e^2[0+0]=0$; direct check $1^2e^2=e^2$ ✓.

Verification, $z=0.5-1.2i$

Numerical substitution gives $w\approx0.18197+4.59029i$ by direct computation, matching the formula for $u,v$ above to full precision.

Result

$$ u(x,y)=e^{2x}\big[(x^2-y^2)\cos2y-2xy\sin2y\big], \qquad v(x,y)=e^{2x}\big[(x^2-y^2)\sin2y+2xy\cos2y\big] $$

A routine application of the multiplication rule for complex numbers once each factor is separately split into real and imaginary parts.