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If z=4-3i write z squared + 17 in the form a+bi, a,b E R.. Hence solve k(z2+17)=|z|(1-i)

Sagot :

konrad509
[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} [/tex]
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