If [tex]z = 4 - 3i[/tex], write [tex]z^2 + 17[/tex] in the form [tex]a + bi[/tex], where [tex]a, b \in \mathbb{R}[/tex].

Hence, solve [tex]k(z^2 + 17) = |z|(1 - i)[/tex].

[z^2+17=(4-3i)^2+17=16-24i-9+17=24-24i\ |z|=\sqrt{4^2+(-3)^2}=\sqrt{16+9}=\sqrt{25}=5\\ k(z^2+17)=|x|(1-i)\ k(24-24i)=5(1-i)\ 24k(1-i)=5(1-i)\ 24k=5\ k=\frac{5}{24}
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Answered by konrad509 | 2024-06-10

We computed z 2 + 17 as 24 − 24 i and found the modulus of z to be 5 . By solving the equation k ( z 2 + 17 ) = ∣ z ∣ ( 1 − i ) , we determined that k = 24 5 ​ .
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Answered by konrad509 | 2024-12-26