Example 8: exam-style question

The complex number

{z}

is given by

{z=3+bi}

, where

{b}

is a real number.

(a) Find the possible values of

{b}

if

\frac{z^2}{z^*}

is real.

For the rest of the question, it is further given that

{b>0}

.

(b) Find the smallest integer value of

{n}

such that

{\left | z^n \right | > 1000}

.

(c) For the value of

{n}

found in (b), find the values of

{\left | z^n \right |}

and

{\arg\left(z^n\right)}

, where

{-\pi <a \arg \left ( z^n \right) \leq \pi}

.

(d) On a single Argand diagram mark out the points

{A,B,C,D}

and

{E}

representing the complex numbers

{z}

,

{\frac{z^2}{z^*}}

,

{z^*}

,

{\frac{18}{z}}

and

{\frac{z^2}{6}}

respectively.

We first evaluate

{\frac{z^2}{z^*}}

in terms of

{b}

\begin{aligned} \frac{z^2}{z^*} &= \frac{\left (3+bi \right)^2}{3-bi} \\ &= \frac{9-b^2+6bi}{3-bi} \cdot \frac{3+bi}{3+bi} \\ &= \frac{3(9-b^2)-6b^2+\left( 18b+(9-b^2)b \right)i}{9+b^2} \end{aligned}

Since

{\frac{z^2}{z^*}}

is real,

{18b+(9-b^2)b=0}

{b(27-b^2)=0}

{b=0, b=\pm 3\sqrt{3}}

{b=3\sqrt{3}}

since

{b>0}

\begin{aligned} \left | z^n \right | &> 1000 \\ \left | 3 + 3\sqrt{3}i \right |^n &> 1000 \\ \left ( \sqrt{3^2 + (3\sqrt{3})^2} \right )^n &> 1000 \\ n \ln 6 &> \ln 1000 \\ n &> 3.8553 \end{aligned}

Smallest integer value of

{n=4}

{\left |z^n \right | = 6^4 = 1296}

{\arg \left ( z \right ) = \tan^{-1} \left ( \frac{3\sqrt{3}}{3} \right) = \frac{\pi}{3}}

{\arg \left ( z^n \right ) = 4 \arg (z) \equiv \frac{4\pi}{3}-2\pi = -\frac{2\pi}{3} }

{z=6e^{i\frac{\pi}{3}}, \quad \frac{z^2}{z^*} = \frac{36e^{i\frac{2\pi}{3}}}{6e^{-i\frac{\pi}{3}}} = 6e^{i\pi}, \quad z^* = 6e^{-i\frac{\pi}{3}}, }

{\frac{18}{z} = \frac{18}{6e^{i\frac{\pi}{3}}} = 3e^{-i\frac{\pi}{3}}, \quad \frac{z^2}{6} = \frac{36e^{i\frac{2\pi}{3}}}{6} = 6e^{i\frac{2\pi}{3}} }

Notes