Problem 31
Medium Algebra of Complex Numbers

Knowing that arg(1+z) = (π)/(4) and |1+z| = 4 , determine…

Knowing that $\arg(1+z) = \dfrac{\pi}{4}$ and $|1+z| = 4$, determine the complex number $z$.

Explanation of Method

The two given facts pin down $1+z$ completely in polar form ($r=4$, $\theta=\frac\pi4$), so convert directly to Cartesian form and subtract $1$ to isolate $z$.

Step 1 · Write $1+z$ in polar form

$$ 1+z = 4e^{i\pi/4} = 4\left(\cos\frac\pi4+i\sin\frac\pi4\right) $$

Step 2 · Convert to Cartesian form

Using $\cos\frac\pi4=\sin\frac\pi4=\dfrac{\sqrt2}{2}$:

$$ 1+z = 4\left(\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}\right) = 2\sqrt2+2\sqrt2\,i $$

Step 3 · Solve for $z$

$$ z = (2\sqrt2+2\sqrt2\,i) - 1 = \boxed{(2\sqrt2-1)+2\sqrt2\,i} $$

Step 4 · Verify

$$ 1+z = 2\sqrt2+2\sqrt2\,i, \qquad |1+z| = \sqrt{(2\sqrt2)^2+(2\sqrt2)^2} = \sqrt{8+8} = \sqrt{16}=4 \quad\checkmark $$

Since the real and imaginary parts of $1+z$ are equal and positive, $\arg(1+z)=\arctan(1)=\dfrac\pi4$. ✓

Step 5 · Cross-check via direct Cartesian substitution

Let $z=x+iy$, so $1+z=(1+x)+iy$. From $|1+z|=4$: $(1+x)^2+y^2=16$. From $\arg(1+z)=\frac\pi4$: $\tan\frac\pi4=\dfrac{y}{1+x}=1$, so $y=1+x$. Substituting:

$$ (1+x)^2+(1+x)^2=16 \quad\Longrightarrow\quad 2(1+x)^2=16 \quad\Longrightarrow\quad (1+x)^2=8 $$

Since $\arg(1+z)=\frac\pi4$ is in the first quadrant, we need $1+x>0$, so $1+x=2\sqrt2$ (not $-2\sqrt2$), giving $x=2\sqrt2-1$ and $y=1+x=2\sqrt2$ — matching Steps 2–3 exactly.

Step 6 · Numeric check

$$ 2\sqrt2 \approx 2.828 \quad\Longrightarrow\quad z \approx 1.828+2.828i $$

$$ 1+z \approx 2.828+2.828i, \qquad |1+z|\approx\sqrt{2\cdot2.828^2}\approx4.000, \qquad \arg(1+z)=45° \quad\checkmark $$

$1$ $1+z=4e^{i\pi/4}$ $z$ Re Im

$1+z$ sits at distance $4$ from the origin at angle $45°$ (dashed, from origin); $z$ is that same point shifted one unit left (solid, from $x=1$).

Result

$$ z = (2\sqrt2-1)+2\sqrt2\,i \approx 1.828+2.828i $$

Equivalently $z = 4e^{i\pi/4}-1$, verified against both the modulus and argument conditions and cross-checked via direct Cartesian substitution.