Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 3.0 AMC 10/12 Find the answer

Given a complex number $Z_1 = 2 + ai$ (where $a \in \mathbb{R}$ and $a>0$ , $i$ is the imaginary unit), and $Z_1^2$ is a pure imaginary number.
(1) Find the value of the real number $a$ .
(2) If $Z = \frac{Z_1}{1-i}$ , find the modulus of the complex number $Z$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

(1) From $Z_1 = 2 + ai$ ,
we have $Z_1^2 = (2+ai)^2 = 4 - a^2 + 4ai$ .
Since $Z_1^2$ is a pure imaginary number and $a > 0$ , we have the system of equations:
$\begin{cases} 4 - a^2 = 0,\\ 4a \neq 0. \end{cases}$
Solving for $a$ , we get $a = 2$ .

(2) To find $Z = \frac{Z_1}{1-i}$ , substitute the value found for $Z_1$ ,
$Z = \frac{2 + 2i}{1 - i} = \frac{(2 + 2i)(1 + i)}{(1 - i)(1 + i)} = \frac{2 + 2i + 2i - 2}{1 + 1} = \frac{4i}{2} = 2i.$
The modulus of $Z$ is the absolute value of a complex number, calculated by
$|Z| = |2i| = \boxed{2}.$

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.